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Integration

1.7 Integration by parts

29 problems · hints, answers and solutions shown beside each one

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

The method of integration by substitution comes from the \rule{1.5cm}{0.15mm} rule for differentiation.
The method of integration by parts comes from the \rule{1.5cm}{0.15mm} rule for differentiation.

Hint

Read back over Sections 1.4 and 1.7 of the CLP-2 text. When these methods are introduced, they are justified using the corresponding differentiation rules.

Answer

chain; product

Full solution

Integration by substitution is just using the chain rule, backwards:

ddx{f(g(x))}=f′(g(x))g′(x)⇔∫ddx{f(g(x))}dx=∫f′(g(x))g′(x)dx⇔f(g(x))⏟f(u)+C=∫f′(g(x))⏟f′(u)g′(x)dx⏟du\begin{alignat*}{3}&&\diff{}{x}\{f(g(x))\}&=f'(g(x))g'(x)\\ &\Leftrightarrow& \int\diff{}{x}\{f(g(x))\}\dee{x} &=\int f'(g(x))g'(x)\dee{x}\\ &\Leftrightarrow&\underbrace{ f(g(x))}_{f(u)} +C &=\int \underbrace{f'(g(x))}_{f'(u)}\underbrace{g'(x)\dee{x}}_{\dee{u}}\end{alignat*}

Similarly, integration by parts comes from the product rule:

ddx{f(x)g(x)}=f′(x)g(x)+f(x)g′(x)⇔∫ddx{f(x)g(x)}dx=∫f′(x)g(x)+f(x)g′(x)dx⇔f(x)g(x)+C=∫f′(x)g(x)dx+∫f(x)g′(x)dx⇔∫f(x)⏟ug′(x)dx⏟dv=f(x)⏟ug(x)⏟v−∫g(x)⏟vf′(x)dx⏟du\begin{alignat*}{3}&&\diff{}{x}\{f(x)g(x)\}&=f'(x)g(x)+f(x)g'(x)\\ &\Leftrightarrow&\int\diff{}{x}\{f(x)g(x)\}\dee{x} &=\int f'(x)g(x)+f(x)g'(x)\dee{x}\\ &\Leftrightarrow&f(x)g(x)+C&=\int f'(x)g(x)\dee{x}+\int f(x)g'(x)\dee{x}\\ &\Leftrightarrow&\int\underbrace{ f(x)}_{u}\underbrace{g'(x)\dee{x}}_{\dee{v}}&=\underbrace{f(x)}_{u}\underbrace{g(x)}_{v}-\int \underbrace{g(x)}_{v}\underbrace{f'(x)\dee{x}}_{\dee{u}}\end{alignat*}

In the last line, the “+C+C” has been absorbed into the indefinite integral on the right hand side.

Q2Stage 1

Suppose you want to evaluate an integral using integration by parts. You choose part of your integrand to be uu, and part to be dv\dee{v}. The part chosen as uu will be: (differentiated, antidifferentiated). The part chosen as dv\dee{v} will be: (differentiated, antidifferentiated).

Hint

Remember our rule: ∫udv=uv−∫vdu\int u \dee{v} = uv - \int v \dee{u}. So, we take uu and use it to make du\dee{u}, and we take dv\dee{v} and use it to make vv.

Answer

The part chosen as uu will be differentiated. The part chosen as dv\dee{v} will be antidifferentiated.

Full solution

Remember our rule: ∫udv=uv−∫vdu\int u \dee{v} = uv - \int v \dee{u}. So, we take uu and use it to make du\dee{u}–that is, we differentiate it. We take dv\dee{v} and use it to make vv–that is, we antidifferentiate it.

Q3Stage 1

Let f(x)f(x) and g(x)g(x) be differentiable functions. Using the quotient rule for differentiation, give an equivalent expression to ∫f′(x)g(x) dx\displaystyle\int \frac{f'(x)}{g(x)}~\dee{x}.

Hint

According to the quotient rule,

ddx{f(x)g(x)}=g(x)f′(x)−f(x)g′(x)g2(x).\diff{}{x}\left\{\dfrac{f(x)}{g(x)}\right\} = \frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}.

Antidifferentiate both sides of the equation, then solve for the expression in the question.

Answer

∫f′(x)g(x) dx=f(x)g(x)+∫f(x)g′(x)g2(x)dx\displaystyle\int \frac{f'(x)}{g(x)}~\dee{x}= \dfrac{f(x)}{g(x)} +\displaystyle\int\frac{f(x)g'(x)}{g^2(x)}\dee{x}

Full solution

We'll use the same ideas that led to the method of integration by parts. (You can review this in your text, or see the solution to Question 1 in this section.) According to the quotient rule,

ddx{f(x)g(x)}=g(x)f′(x)−f(x)g′(x)g2(x).\begin{align*}\diff{}{x}\left\{\dfrac{f(x)}{g(x)}\right\} &= \frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}.\end{align*}

Antidifferentiating both sides gives us:

∫ddx{f(x)g(x)}dx=∫g(x)f′(x)−f(x)g′(x)g2(x)dxf(x)g(x)+C=∫f′(x)g(x)dx−∫f(x)g′(x)g2(x)dx∫f′(x)g(x)dx=f(x)g(x)+∫f(x)g′(x)g2(x)dx\begin{align*}\int\diff{}{x}\left\{\dfrac{f(x)}{g(x)}\right\} \dee{x}&= \int\frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}\dee{x}\\ \dfrac{f(x)}{g(x)} +C&= \int\frac{f'(x)}{g(x)}\dee{x}-\int\frac{f(x)g'(x)}{g^2(x)}\dee{x} \\\int\frac{f'(x)}{g(x)}\dee{x} &= \dfrac{f(x)}{g(x)}+\int\frac{f(x)g'(x)}{g^2(x)}\dee{x}\end{align*}

In the last line, the “+C+C” has been absorbed into the indefinite integral on the right hand side.

This is exactly the integration by parts formula for the functions u=1/gu=1/g and v=fv=f.

Q4Stage 1

Suppose we want to use integration by parts to evaluate ∫u(x)⋅v′(x)dx\displaystyle\int u(x)\cdot v'(x) \dee{x} for some differentiable functions uu and vv. We need to find an antiderivative of v′(x)v'(x), but there are infinitely many choices. Show that every antiderivative of v′(x)v'(x) gives an equivalent final answer.

Hint

Remember all the antiderivatives differ only by a constant, so you can write them all as v(x)+Cv(x)+C for some CC.

Answer

All the antiderivatives differ only by a constant, so we can write them all as v(x)+Cv(x)+C for some CC. Then, using the formula for integration by parts,

∫u(x)⋅v′(x)dx=u(x)⏟u[v(x)+C]⏟v−∫[v(x)+C]⏟vu′(x)dx⏟du=u(x)v(x)+Cu(x)−∫v(x)u′(x)dx−∫Cu′(x)dx=u(x)v(x)+Cu(x)−∫v(x)u′(x)dx−Cu(x)+D=u(x)v(x)−∫v(x)u′(x)dx+D\begin{align*} \int u(x)\cdot v'(x) \dee{x}&=\underbrace{u(x)}_u\underbrace{\big[ v(x)+C\big]}_v - \int \underbrace{\big[ v(x)+C\big]}_v \underbrace{u'(x)\dee{x}}_{\dee{u}}\\ &=u(x)v(x)+Cu(x) - \int v(x)u'(x)\dee{x} - \int Cu'(x)\dee{x} \\&=u(x)v(x)+Cu(x) - \int v(x)u'(x)\dee{x} - Cu(x)+D \\&=u(x)v(x) - \int v(x)u'(x)\dee{x} +D \end{align*}

where DD is any constant.

Since the terms with CC cancel out, it didn't matter what we chose for CC–all choices end up the same.

Full solution

All the antiderivatives differ only by a constant, so we can write them all as v(x)+Cv(x)+C for some CC. Then, using the formula for integration by parts,

∫u(x)⋅v′(x)dx=u(x)⏟u[v(x)+C]⏟v−∫[v(x)+C]⏟vu′(x)dx⏟du=u(x)v(x)+Cu(x)−∫v(x)u′(x)dx−∫Cu′(x)dx=u(x)v(x)+Cu(x)−∫v(x)u′(x)dx−Cu(x)+D=u(x)v(x)−∫v(x)u′(x)dx+D\begin{align*} \int u(x)\cdot v'(x) \dee{x}&=\underbrace{u(x)}_u\underbrace{\big[ v(x)+C\big]}_v - \int \underbrace{\big[ v(x)+C\big]}_v \underbrace{u'(x)\dee{x}}_{\dee{u}}\\ &=u(x)v(x)+Cu(x) - \int v(x)u'(x)\dee{x} - \int Cu'(x)\dee{x} \\&=u(x)v(x)+Cu(x) - \int v(x)u'(x)\dee{x} - Cu(x)+D \\&=u(x)v(x) - \int v(x)u'(x)\dee{x} +D \end{align*}

where DD is any constant.

Since the terms with CC cancel out, it didn't matter what we chose for CC–all choices end up the same.

Q5Stage 1

Suppose you want to evaluate ∫f(x)dx\displaystyle\int f(x)\dee{x} using integration by parts. Explain why dv=f(x)dx\dee{v} = f(x)\dee{x}, u=1u=1 is generally a bad choice.

Note: compare this to Example 1.7.8 of the CLP-2 text, where we chose u=f(x)u=f(x), dv=1dx\dee{v}=1\dee{x}.

Hint

What integral do you have to evaluate, after you plug in your choices to the integration by parts formula?

Answer

Suppose we choose dv=f(x)dx\dee{v} = f(x)\dee{x}, u=1u=1. Then v=∫f(x)dxv = \displaystyle\int f(x)\dee{x}, and du=dx\dee{u}=\dee{x}. So, our integral becomes:

∫(1)⏟uf(x)dx⏟dv=(1)⏟u∫f(x)dx⏟v−∫(∫f(x)dx)⏟vdx⏟du\begin{align*} \int \underbrace{(1)}_{u}\underbrace{f(x)\dee{x}}_{\dee{v}}&= \underbrace{(1)}_{u}\underbrace{\int f(x)\dee{x}}_{v} - \int\underbrace{ \left(\int f(x)\dee{x}\right)}_{v}\underbrace{\dee{x}}_{\dee{u}} \end{align*}

In order to figure out the first product (and the second integrand), you need to know the antiderivative of f(x)f(x)–but that's exactly what you're trying to figure out!

Full solution

Suppose we choose dv=f(x)dx\dee{v} = f(x)\dee{x}, u=1u=1. Then v=∫f(x)dxv = \displaystyle\int f(x)\dee{x}, and du=dx\dee{u}=\dee{x}. So, our integral becomes:

∫(1)⏟uf(x)dx⏟dv=(1)⏟u∫f(x)dx⏟v−∫(∫f(x)dx)⏟vdx⏟du\begin{align*} \int \underbrace{(1)}_{u}\underbrace{f(x)\dee{x}}_{\dee{v}}&= \underbrace{(1)}_{u}\underbrace{\int f(x)\dee{x}}_{v} - \int\underbrace{ \left(\int f(x)\dee{x}\right)}_{v}\underbrace{\dee{x}}_{\dee{u}} \end{align*}

In order to figure out the first product (and the second integrand), you need to know the antiderivative of f(x)f(x)–but that's exactly what you're trying to figure out! So, using integration by parts has not eased your pain.

We note here that in certain cases, such as ∫log⁡x dx\int\log x~ \dee{x} (Example 1.7.8 in the CLP-2 text), it is useful to choose dv=1dx\dee{v}=1\dee{x}. This is similar to, but crucially different from, the do-nothing method in this problem.

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q6Stage 2Past exam · M121 2014A

Evaluate ∫xlog⁡x dx{\displaystyle\int x\log x\,\dee{x}}.

Hint

You'll probably want to use integration by parts. (It's the title of the section, after all). You'll break the integrand into two parts, integrate one, and differentiate the other. Would you rather integrate log⁡x\log x, or differentiate it?

Answer

x2log⁡x2−x24+C\dfrac{x^2\log x}{2} - \dfrac{x^2}{4} + C

Full solution

For integration by parts, we want to break the integrand into two pieces, multiplied together. There is an obvious choice for how to do this: one piece is xx, and the other is log⁡x\log x. Remember that one piece will be integrated, while the other is differentiated. The question is, which choice will be more helpful. After some practice, you'll get the hang of making the choice. For now, we'll present both choices–but when you're writing a solution, you don't have to write this part down. It's enough to present your choice, and then a successful computation is justification enough.

Option 1:u=xdu=1 dxdv=log⁡x dxv=??Option 2:u=12x2du=log⁡x dxdv=1x dxv=??\begin{array}{c|l|l|}\hline \text{Option 1:}& u=x & \mathrm{d}u=1\,\mathrm{d}x\\ & \mathrm{d}v=\log x\,\mathrm{d}x & v=??\\\hline \text{Option 2:}& u=\dfrac{1}{2}x^2 & \mathrm{d}u=\log x\,\mathrm{d}x\\ & \mathrm{d}v=\dfrac{1}{x}\,\mathrm{d}x & v=??\\\hline\end{array}

In Example 1.7.8 of the CLP-2 text, we found the antiderivative of logarithm, but it wasn't trivial. We might reasonably want to avoid using this complicated antiderivative. Indeed, Option 2 (differentiating logarithm, antidifferentiating xx) looks promising–when we multiply the blue equations, we get something easily integrable– so let's not even bother going deeper into Option 1.

That is, we perform integration by parts with u=log⁡xu=\log x and dv=x dx\dee{v}=x\,\dee{x}, so that du=dxx\dee{u}=\frac{\dee{x}}{x} and v=x22v = \frac{x^2}{2}.

∫log⁡x⏟ux dx⏟dv=x2log⁡x2⏟uv−∫x22⏟v dxx⏟du=x2log⁡x2−12∫x dx=x2log⁡x2−x24+C\begin{align*} \int\underbrace{\log x}_{u}\underbrace{x\,\dee{x}}_{\dee{v}} &= \underbrace{\frac{x^2\log x}{2}}_{uv} - \int \underbrace{\frac{x^2}{2}}_v\ \underbrace{\frac{\dee{x}}{x}}_{\dee{u}} = \frac{x^2\log x}{2} -\frac{1}{2} \int x \ \dee{x} \\ &= \frac{x^2\log x}{2} - \frac{x^2}{4} + C \end{align*}
Q7Stage 2Past exam · M105 2013A

Evaluate ∫log⁡xx7 dx{\displaystyle\int \frac{\log x}{x^7}\,\dee{x}}.

Hint

This problem is similar to Question 6.

Answer

−log⁡x6x6−136x6+C- \dfrac{\log x}{6 x^6} - \dfrac{1}{36 x^6} + C

Full solution

Our integrand is the product of two functions, and there's no clear substitution. So, we might reasonably want to try integration by parts. Again, we have two obvious pieces: log⁡x\log x, and x−7x^{-7}. We'll consider our options for assigning these to uu and dv\dee{v}:

Option 1:u=log⁡xdu=1x dxdv=x−7 dxv=1−6x−6Option 2:u=??du=x−7 dxdv=−7x−8 dxv=1−6x−6\begin{array}{c|l|l|}\hline \text{Option 1:}& u=\log x & \mathrm{d}u=\dfrac{1}{x}\,\mathrm{d}x\\ & \mathrm{d}v=x^{-7}\,\mathrm{d}x & v=\dfrac{1}{-6}x^{-6}\\\hline \text{Option 2:}& u=?? & \mathrm{d}u=x^{-7}\,\mathrm{d}x\\ & \mathrm{d}v=-7x^{-8}\,\mathrm{d}x & v=\dfrac{1}{-6}x^{-6}\\\hline\end{array}

Again, we remember that logarithm has some antiderivative we found in Example 1.7.8 of the CLP-2 text, but it was something complicated. Luckily, we don't need to bother with it: when we multiply the red equations in Option 1, we get a perfectly workable integral.

We perform integration by parts with u=log⁡xu=\log x and dv=x−7 dx\dee{v}=x^{-7}\,\dee{x}, so that du=dxx\dee{u}=\frac{\dee{x}}{x} and v=−x−66v = -\frac{x^{-6}}{6}.

∫log⁡xx7 dx=−log⁡x x−66⏟uv+∫x−66⏟−v dxx⏟du=−log⁡x6x6+16∫x−7 dx=−log⁡x6x6−136x6+C\begin{align*} \int \frac{\log x}{x^7}\,\dee{x} &=\underbrace{ -\log x\ \frac{x^{-6}}{6}}_{uv} + \int \underbrace{\frac{x^{-6}}{6}}_{-v}\ \underbrace{\frac{\dee{x}}{x}}_{\dee{u}} = - \frac{\log x}{6 x^6} +\frac{1}{6} \int x^{-7} \ \dee{x} \\ &= - \frac{\log x}{6 x^6} - \frac{1}{36 x^6} + C \end{align*}
Q8Stage 2Past exam · 2016A

Evaluate ∫0πxsin⁡x dx\displaystyle\int_0^\pi x\sin x\,\dee{x}.

Hint

Example 1.7.5 in the

CLP-2 text shows you how to find the antiderivative. Then the Fundamental Theorem of Calculus Part 2 gives you the definite integral.

Answer

π\pi

Full solution

To integrate by parts, we need to decide what to use as uu, and what to use as dv\dee{v}. The salient parts of this integrand are xx and sin⁡x\sin x, so we only need to decide which is uu and which dv\dee{v}. Again, this process will soon become familiar, but to help you along we show you both options below.

Option 1:u=xdu=1 dxdv=sin⁡x dxv=−cos⁡xOption 2:u=12x2du=sin⁡x dxdv=cos⁡x dxv=−cos⁡x\begin{array}{c|l|l|}\hline \text{Option 1:}& u=x & \mathrm{d}u=1\,\mathrm{d}x\\ & \mathrm{d}v=\sin x\,\mathrm{d}x & v=-\cos x\\\hline \text{Option 2:}& u=\dfrac{1}{2}x^2 & \mathrm{d}u=\sin x\,\mathrm{d}x\\ & \mathrm{d}v=\cos x\,\mathrm{d}x & v=-\cos x\\\hline\end{array}

The derivative and antiderivative of sine are almost the same, but xx turns into something simpler when we differentiate it. So, we choose Option 1.

We integrate by parts, using u=xu = x, dv=sin⁡x dx\dee{v} =\sin x \,\dee{x} so that v=−cos⁡xv=-\cos x and du=dx\dee{u} = \dee{x}:

∫0πxsin⁡x dx=−xcos⁡x⏟uv∣0π−∫0π(−cos⁡x)⏟v dx⏟du=[−xcos⁡x+sin⁡x]0π=−π(−1)=π\begin{align*} \int_0^\pi x\sin x\,\dee{x} = \underbrace{-x\cos x}_{uv} \Big|_0^\pi -\int_0^\pi \underbrace{(-\cos x)}_v\,\underbrace{\dee{x}}_{\dee{u}} = \Big[-x\cos x +\sin x\Big]_0^\pi =-\pi(-1) =\pi \end{align*}
Q9Stage 2Past exam · M105 2015A

Evaluate ∫0π2xcos⁡x dx\displaystyle\int_0^{\frac{\pi}{2}} x\cos x\,\dee{x}.

Hint

Compare to Question 8. Try to do this one all the way through without peeking at another solution!

Answer

π2−1\dfrac{\pi}{2} -1

Full solution

When we have two functions multiplied like this, and there's no obvious substitution, our minds turn to integration by parts. We hope that our integral will be improved by differentiating one part and antidifferentiating the other. Let's see what our choices are:

Option 1:u=xdu=1 dxdv=cos⁡x dxv=sin⁡xOption 2:u=12x2du=cos⁡x dxdv=−sin⁡x dxv=sin⁡x\begin{array}{c|l|l|}\hline \text{Option 1:}& u=x & \mathrm{d}u=1\,\mathrm{d}x\\ & \mathrm{d}v=\cos x\,\mathrm{d}x & v=\sin x\\\hline \text{Option 2:}& u=\dfrac{1}{2}x^2 & \mathrm{d}u=\cos x\,\mathrm{d}x\\ & \mathrm{d}v=-\sin x\,\mathrm{d}x & v=\sin x\\\hline\end{array}

Option 1 seems preferable. We integrate by parts, using u=xu = x, dv=cos⁡x dx\dee{v} =\cos x \,\dee{x} so that v=sin⁡xv=\sin x and du=dx\dee{u} = \dee{x}:

∫0π2xcos⁡x dx=x⏟usin⁡x⏟v∣0π2−∫0π2sin⁡x⏟v dx⏟du=[xsin⁡x+cos⁡x]0π2=π2−1\begin{align*} \int_0^{\frac{\pi}{2}} x\cos x\,\dee{x} = \underbrace{x}_u\underbrace{\sin x}_v \Big|_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}} \underbrace{\sin x}_v\,\underbrace{\dee{x}}_{\dee{u}} = \Big[x\sin x +\cos x\Big]_0^{\frac{\pi}{2}} =\frac{\pi}{2} - 1 \end{align*}
Q10Stage 2

Evaluate ∫x3exdx\displaystyle\int x^3 e^x \dee{x}.

Hint

If at first you don't succeed, try using integration by parts a few times in a row. Eventually, one part will go away.

Answer

ex(x3−3x2+6x−6)+Ce^x\left(x^3-3x^2+6x - 6\right)+C

Full solution

This integrand is the product of two functions, with no obvious substitution. So, let's try integration by parts, with one part exe^x and one part x3x^3.

Option 1:u=exdu=ex dxdv=x3 dxv=14x4Option 2:u=exdu=x3 dxdv=3x2 dxv=14x4\begin{array}{c|l|l|}\hline \text{Option 1:}& u=e^x & \mathrm{d}u=e^x\,\mathrm{d}x\\ & \mathrm{d}v=x^3\,\mathrm{d}x & v=\frac{1}{4}x^4\\\hline \text{Option 2:}& u=e^x & \mathrm{d}u=x^3\,\mathrm{d}x\\ & \mathrm{d}v=3x^2\,\mathrm{d}x & v=\frac{1}{4}x^4\\\hline\end{array}

At first glance, multiplying the red functions and multiplying the blue functions give largely equivalent integrands to what we started with–none of them with obvious antiderivatives. In previous questions, we were able to choose u=xu=x, and then du=dx\dee{u}=\dee{x}, so the “xx" in the integrand effectively went away. Here, we see that choosing u=x3u=x^3 will lead to du=3x2dx\dee{u}=3x^2\dee{x}, which has a lower power. If we repeatedly perform integration by parts, choosing uu to be the power of xx each time, then after a few iterations it should go away, because the third derivative of x3x^3 is a constant.

So, we start with Option 2: u=x3u=x^3, dv=exdx\dee{v}=e^x\dee{x}, du=3x2dx\dee{u}=3x^2\dee{x}, and v=exv = e^x.

∫x3exdx=x3⏟uex⏟v−∫ex⏟v⋅3x2dx⏟du=x3ex−3∫ex⋅x2dx\begin{align*}\int x^3 e^x \dee{x}&= \underbrace{x^3}_{u}\underbrace{e^x}_{v} - \int \underbrace{e^x}_v \cdot\underbrace{ 3x^2 \dee{x}}_{\dee{u}}\\ &= {x^3}{e^x} -3 \int {e^x} \cdot x^2 \dee{x}\end{align*}

Now, we take u=x2u=x^2 and dv=exdx\dee{v}=e^x\dee{x}, so du=2xdx\dee{u}=2x\dee{x} and v=exv=e^x. We're only using integration by parts on the actual integral–the rest of the function stays the way it is.

=x3ex−3[x2ex⏟uv−∫ex⏟v⋅2xdx⏟du]=x3ex−3x2ex+6∫xexdx\begin{align*}&=x^3e^x -3 \left[\underbrace{x^2e^x}_{uv} - \int \underbrace{e^x}_{v} \cdot \underbrace{2x\dee{x}}_{\dee{u}}\right] \\ &=x^3e^x - 3x^2e^x + 6\int x e^x\dee{x}\end{align*}

Continuing, we take u=xu=x and dv=exdx\dee{v}=e^x\dee{x}, so du=dx\dee{u}=\dee{x} and v=exv=e^x. This is the step where the polynomial part of the integrand finally disappears.

=x3ex−3x2ex+6[xex⏟uv−∫ex⏟vdx⏟du]=x3ex−3x2ex+6xex−6ex+C=ex(x3−3x2+6x−6)+C\begin{align*}&=x^3e^x - 3x^2e^x + 6\left[\underbrace{xe^x}_{uv} - \int \underbrace{ e^x}_v \underbrace{\dee{x}}_{\dee{u}}\right]\\ &=x^3e^x-3x^2e^x+6xe^x - 6e^x+C \\ &=e^x\left(x^3-3x^2+6x - 6\right)+C\end{align*}

Let's check that this makes sense: the derivative of ex(x3−3x2+6x−6)+Ce^x\left(x^3-3x^2+6x - 6\right)+C should be x3exx^3e^x. We differentiate using the product rule.

ddx{ex(x3−3x2+6x−6)+C}=ex(x3−3x2+6x−6)+ex(3x2−6x+6)=ex(x3−3x2+3x2+6x−6x−6+6)=x3ex\begin{align*} \diff{}{x}\left\{e^x\left(x^3-3x^2+6x - 6\right)+C\right\} &=e^x\left(x^3-3x^2+6x - 6\right)+e^x\left(3x^2-6x+6\right)\\ &=e^x\left(x^3-3x^2+3x^2+6x-6x-6+6\right)=x^3e^x \end{align*}

Remark: In order to be technically correct in our antidifferentiation, we should add the +C+C as soon as we do the first integration by parts. However, when we are using integration by parts, we usually end up evaluating an integral at the end, and we add the +C+C at that point. Since the +C+C comes up eventually, it is common practice to not clutter our calculations with it until the end.

Q11Stage 2

Evaluate ∫xlog⁡3x dx\displaystyle\int x \log^3 x~ \dee{x}.

Hint

Similarly to Question 10, look for a way to use integration by parts a few times to simplify the integrand until it is antidifferentiatable.

Answer

x22log⁡3x−3x24log⁡2x+3x24log⁡x−3x28+C\dfrac{x^2}{2}\log^3x - \dfrac{3x^2}{4}\log^2 x + \dfrac{3x^2}{4}\log x - \dfrac{3x^2}{8}+C

Full solution

Since our integrand is two functions multiplied together, and there isn't an obvious substitution, let's try integration by parts. Here are our salient options.

Option 1:u=xdu=1 dxdv=log⁡3x dxv=??Option 2:u=12x2du=log⁡3x dxdv=3log⁡2x⋅1x dxv=??\begin{array}{c|l|l|}\hline \text{Option 1:}& u=x & \mathrm{d}u=1\,\mathrm{d}x\\ & \mathrm{d}v=\log^3 x\,\mathrm{d}x & v=??\\\hline \text{Option 2:}& u=\dfrac{1}{2}x^2 & \mathrm{d}u=\log^3 x\,\mathrm{d}x\\ & \mathrm{d}v=3\log^2 x \cdot\frac{1}{x}\,\mathrm{d}x & v=??\\\hline\end{array}

This calls for some strategizing. Using the template of Example 1.7.8 in the CLP-2 text, we could probably figure out the antiderivative of log⁡3x\log^3 x. Option 1 is tempting, because our xx-term goes away. So, there might be a benefit there, but on the other hand, the antiderivative of log⁡3x\log^3 x is probably pretty complicated.

Now let's consider Option 2. When we multiply the blue functions together, we get something similar to our original integrand, but the power of logarithm is smaller. If we were to iterate this method (using integration by parts a few times, always choosing uu to be the part with a logarithm) then eventually we would end up differentiating logarithm. This seems like a safer plan: let's do Option 2.

We use integration by parts with u=log⁡3xu=\log^3 x, dv=xdx\dee{v} = x\dee{x}, du=3xlog⁡2xdx\dee{u}=\frac{3}{x}\log^2 x \dee{x}, and v=12x2v = \frac{1}{2}x^2.

∫xlog⁡3x dx=12x2log⁡3x⏟uv−∫32xlog⁡2xdx⏟vdu=12x2log⁡3x−32∫xlog⁡2xdx\begin{align*}\int x \log^3 x~ \dee{x}&=\underbrace{\frac{1}{2}x^2\log^3 x}_{uv} - \int\underbrace{ \frac{3}{2}x\log^2 x \dee{x}}_{v\dee{u}}\\ &=\frac{1}{2}x^2\log^3 x- \frac{3}{2}\int x\log^2 x \dee{x}\end{align*}

Continuing our quest to differentiate away the logarithm, we use integration by parts with u=log⁡2xu=\log^2x, dv=xdx\dee{v} = x\dee{x}, du=2xlog⁡xdx\dee{u} = \dfrac{2}{x}\log x\dee{x}, and v=12x2v = \dfrac{1}{2}x^2.

=12x2log⁡3x−32[12x2log⁡2x⏟uv−∫xlog⁡xdx⏟vdu]=12x2log⁡3x−34x2log⁡2x+32∫xlog⁡xdx\begin{align*}&=\frac{1}{2}x^2\log^3x - \frac{3}{2}\left[ \underbrace{\frac{1}{2}x^2\log^2 x}_{uv} - \int \underbrace{x\log x \dee{x}}_{v\dee{u}} \right]\\ &=\frac{1}{2}x^2\log^3x - \frac{3}{4}x^2\log^2 x + \frac{3}{2}\int x\log x \dee{x}\end{align*}

One last integration by parts: u=log⁡xu=\log x, dv=xdx\dee{v}=x\dee{x}, du=1xdx\dee{u}=\dfrac{1}{x}\dee{x}, and v=12x2v = \dfrac{1}{2}x^2.

=12x2log⁡3x−34x2log⁡2x+32[12x2log⁡x⏟uv−∫12xdx⏟vdu]=12x2log⁡3x−34x2log⁡2x+34x2log⁡x−34∫xdx=12x2log⁡3x−34x2log⁡2x+34x2log⁡x−38x2+C\begin{align*}&=\frac{1}{2}x^2\log^3x - \frac{3}{4}x^2\log^2 x + \frac{3}{2}\left[ \underbrace{\frac{1}{2}x^2\log x}_{uv} - \int \underbrace{\frac{1}{2}x\dee{x}}_{v\dee{u}} \right]\\ &=\frac{1}{2}x^2\log^3x - \frac{3}{4}x^2\log^2 x + \frac{3}{4}x^2\log x - \frac{3}{4}\int x\dee{x}\\ &=\frac{1}{2}x^2\log^3x - \frac{3}{4}x^2\log^2 x + \frac{3}{4}x^2\log x - \frac{3}{8}x^2+C\end{align*}

Once again, technically there is a +C+C in the work after the first integration by parts, but we follow convention by conveniently suppressing it until the final integration.

Q12Stage 2

Evaluate ∫x2sin⁡x dx\displaystyle\int x^2\sin x~\dee{x}.

Hint

Use integration by parts twice to get an integral with only a trigonometric function in it.

Answer

(2−x2)cos⁡x+2xsin⁡x+C(2-x^2)\cos x + 2x\sin x +C

Full solution

The integrand is the product of two functions, without an obvious substitution, so let's see what integration by parts can do for us.

Option 1:u=x2du=2x dxdv=sin⁡x dxv=−cos⁡xOption 2:u=13x3du=sin⁡x dxdv=cos⁡x dxv=−cos⁡x\begin{array}{c|l|l|}\hline \text{Option 1:}& u=x^2 & \mathrm{d}u=2x\,\mathrm{d}x\\ & \mathrm{d}v=\sin x\,\mathrm{d}x & v=-\cos x\\\hline \text{Option 2:}& u=\frac{1}{3}x^3 & \mathrm{d}u=\sin x\,\mathrm{d}x\\ & \mathrm{d}v=\cos x\,\mathrm{d}x & v=-\cos x\\\hline\end{array}

Neither option gives us something immediately integrable, but Option 1 replaces our x2x^2 term with a lower power of xx. If we repeatedly apply integration by parts, we can reduce this power to zero. So, we start by choosing u=x2u=x^2 and dv=sin⁡xdx\dee{v}=\sin x\dee{x}, so du=2xdx\dee{u}=2x\dee{x} and v=−cos⁡xv=-\cos x.

∫x2sin⁡x dx=−x2cos⁡x⏟uv+∫2xcos⁡xdx⏟−vdu=−x2cos⁡x+2∫xcos⁡xdx\begin{align*}\int x^2\sin x~\dee{x} &=\underbrace{ -x^2\cos x}_{uv} +\underbrace{ \int 2x\cos x \dee{x}}_{-v\dee{u}}\\ &=-x^2\cos x + 2 \int x \cos x \dee{x}\end{align*}

Using integration by parts again, we want to be differentiating (not antidifferentiating) xx, so we choose u=xu=x, dv=cos⁡xdx\dee{v}=\cos x \dee{x}, and then du=dx\dee{u}=\dee{x} (xx went away!), v=sin⁡xv=\sin x.

=−x2cos⁡x+2[xsin⁡x⏟uv−∫sin⁡xdx⏟vdu]=−x2cos⁡x+2xsin⁡x+2cos⁡x+C=(2−x2)cos⁡x+2xsin⁡x+C\begin{align*}&=-x^2\cos x + 2\left[\underbrace{x\sin x}_{uv}-\int \underbrace{\sin x \dee{x}}_{v\dee{u}} \right] \\&=-x^2\cos x + 2x\sin x+2\cos x +C\\ &=(2-x^2)\cos x + 2x\sin x +C\end{align*}
Q13Stage 2

Evaluate ∫(3t2−5t+6)log⁡t dt\displaystyle\int (3t^2-5t+6)\log t~\dee{t}.

Hint

If you let u=log⁡tu=\log t in the integration by parts, then du\dee{u} works quite nicely with the rest of the integrand.

Answer

(t3−52t2+6t)log⁡t−13t3+54t2−6t+C\left( t^3 - \frac{5}{2}t^2+6t \right)\log t -\frac{1}{3}t^3 +\frac{5}{4}t^2-6t+C

Full solution

This problem is similar to Questions 6 and 7: integrating a polynomial multiplied by a logarithm. Just as in these questions, if we use integration by parts with u=log⁡tu=\log t, then du=1tdt\dee{u} = \dfrac{1}{t}\dee{t}, and our new integrand will consist of powers of tt–which are easy to antidifferentiate.

So, we use u=log⁡tu=\log t, dv=3t2−5t+6\dee{v}= 3t^2-5t+6, du=1tdt\dee{u}=\frac{1}{t}\dee{t}, and v=t3−52t2+6tv = t^3 - \frac{5}{2}t^2+6t.

∫(3t2−5t+6)log⁡t dt=log⁡t⏟u(t3−52t2+6t⏟v)−∫1t(t3−52t2+6t)dt⏟vdu=(t3−52t2+6t)log⁡t−∫(t2−52t+6)dt=(t3−52t2+6t)log⁡t−13t3+54t2−6t+C\begin{align*} \int (3t^2-5t+6)\log t~\dee{t}&=\underbrace{\log t}_u\left(\underbrace{ t^3 - \frac{5}{2}t^2+6t}_v\right) - \int \underbrace{\frac{1}{t}\left( t^3 - \frac{5}{2}t^2+6t\right)\dee{t}}_{v\dee{u}}\\&=\left( t^3 - \frac{5}{2}t^2+6t \right)\log t- \int\left( t^2-\frac{5}{2}t+6 \right)\dee{t}\\&=\left( t^3 - \frac{5}{2}t^2+6t \right)\log t -\frac{1}{3}t^3 +\frac{5}{4}t^2-6t+C \end{align*}
Q14Stage 2

Evaluate ∫sesds\displaystyle\int \sqrt{s}e^{\sqrt{s}}\dee{s}.

Hint

Those square roots are a little disconcerting– get rid of them with a substitution.

Answer

es(2s−4s+4)+Ce^{\sqrt{s}}\left(2s - 4\sqrt{s} +4\right)+C

Full solution

Before we jump to integration by parts, we notice that the square roots lend themselves to substitution. Let's take w=sw=\sqrt{s}. Then dw=12s ds\dee{w}=\dfrac{1}{2\sqrt{s}}~\dee{s}, so 2w dw=ds2w~\dee{w}=\dee{s}.

∫sesds=∫w⋅ew⋅2wdw=2∫w2ewdw\begin{align*}\int \sqrt{s}e^{\sqrt{s}}\dee{s}&=\int w\cdot e^w\cdot 2w\dee{w} = 2\int w^2e^w\dee{w}\end{align*}

Now we have nearly the situation of Question 10. We can repeatedly use integration by parts, with uu as the power of ww, to get rid of the polynomial part. We'll start with u=w2u=w^2, dv=ewdw\dee{v}=e^w\dee{w}, du=2wdw\dee{u}=2w\dee{w}, and v=ewv=e^w.

=2[w2ew⏟uv−∫2wewdw⏟vdu]=2w2ew−4∫wewdw\begin{align*}&=2\left[\underbrace{w^2e^w}_{uv} - \int \underbrace{2we^w\dee{w}}_{v\dee{u}} \right]\\ &=2w^2e^w - 4\int we^w \dee{w}\end{align*}

We use integration by parts again, this time with u=wu=w, dv=ewdw\dee{v}=e^w\dee{w}, du=dw\dee{u}=\dee{w}, and v=ewv=e^w.

=2w2ew−4[wew⏟uv−∫ewdw⏟vdu]=2w2ew−4wew+4ew+C=ew(2w2−4w+4)+C=es(2s−4s+4)+C\begin{align*}&=2w^2e^w - 4\left[\underbrace{we^w}_{uv} - \int\underbrace{ e^w\dee{w}}_{v\dee{u}}\right]\\ &=2w^2e^w - 4we^w +4e^w+C\\ &=e^w\left(2w^2 - 4w +4\right)+C\\ &=e^{\sqrt{s}}\left(2s - 4\sqrt{s} +4\right)+C\end{align*}
Q15Stage 2

Evaluate ∫log⁡2xdx\displaystyle\int \log^2 x \dee{x}.

Hint

This can be solved using the same ideas as Example 1.7.8 in the CLP-2 text.

Answer

xlog⁡2x−2xlog⁡x+2x+Cx\log^2 x -2x\log x +2x+C

Full solution

Let's use integration by parts. What are our parts? We have a few options.

  • Following Example 1.7.8 in the CLP-2 text, we choose u=log⁡2xu=\log^2 x and dv=dx\dee{v}=\dee{x}, so that du=2xlog⁡x dx\dee{u}=\frac{2}{x}\log x ~\dee{x} and v=xv=x.

    ∫log⁡2xdx=xlog⁡2x⏟uv−∫2log⁡xdx⏟vdu\begin{align*}\int \log^2 x \dee{x}&= \underbrace{x\log^2x}_{uv} - \int \underbrace{ 2\log x \dee{x}}_{v\dee{u}}\end{align*}

    Here we can either use the antiderivative of logarithm from memory, or re-derive it. We do the latter, using integration by parts with u=log⁡xu=\log x, dv=2dx\dee{v}=2\dee{x}, du=1xdx\dee{u}=\frac{1}{x}\dee{x}, and v=2xv=2x.

    =xlog⁡2x−[2xlog⁡x⏟uv−∫2dx⏟vdu]=xlog⁡2x−2xlog⁡x+2x+C\begin{align*}&=x\log^2 x - \left[\underbrace{2x\log x}_{uv} - \int \underbrace{2\dee{x}}_{v\dee{u}}\right]\\ &=x\log^2 x - 2x\log x +2x+C\end{align*}

  • Our integrand is two functions multiplied together: log⁡x\log x and log⁡x\log x. So, we will use integration by parts with u=log⁡xu=\log x, dv=log⁡x\dee{v}=\log x, du=1xdx\dee{u}=\frac{1}{x}\dee{x}, and (using the antiderivative of logarithm, found in Example 1.7.8 in the CLP-2 text) v=xlog⁡x−xv=x\log x -x.

    ∫log⁡2x dx=(log⁡x⏟u)(xlog⁡x−x⏟v)−∫(xlog⁡x−x⏟v)1xdx⏟du=xlog⁡2x−xlog⁡x−∫(log⁡x−1)dx=xlog⁡2x−xlog⁡x−[(xlog⁡x−x)−x]+C=xlog⁡2x−2xlog⁡x+2x+C\begin{align*} \int \log^2 x ~\dee{x}&=(\underbrace{\log x}_u) (\underbrace{x\log x - x}_{v}) -\int (\underbrace{x\log x-x}_v)\underbrace{ \frac{1}{x}\dee{x}}_{\dee{u}}\\ &= x\log^2 x - x\log x - \int \left(\log x -1\right)\dee{x}\\ &= x\log^2 x - x\log x - \left[(x\log x - x )-x\right]+C\\ &=x\log^2 x -2x\log x +2x+C \end{align*}
Q16Stage 2

Evaluate ∫2xex2+1dx\displaystyle\int 2xe^{x^2+1}\dee{x}.

Hint

Not every integral should be evaluated using integration by parts.

Answer

ex2+1+Ce^{x^2+1}+C

Full solution

This is your friendly reminder that to a person with a hammer, everything looks like a nail. The integral in the problem is a classic example of an integral to solve using substitution. We have an “inside function," x2+1x^2+1, whose derivative shows up multiplied to the rest of the integrand. We take u=x2+1u=x^2+1, then du=2xdx\dee{u}=2x\dee{x}, so

∫2xex2+1dx=∫eudu=eu+C=ex2+1+C\int 2xe^{x^2+1}\dee{x} = \int e^u\dee{u}=e^u+C = e^{x^2+1}+C
Q17Stage 2Past exam · M105 2015A

Evaluate ∫arccos⁡y dy\displaystyle\int\arccos y\,\dee{y}.

Hint

You know, or can easily look up, the derivative of arccosine. You can use a similar trick as the book did when antidifferentiating other inverse trigonometric functions in Example 1.7.9 of the CLP-2 text.

Answer

yarccos⁡y−1−y2+Cy \arccos y - \sqrt{1-y^2} + C

Full solution

In Example 1.7.9 of the CLP-2 text, we saw that integration by parts was useful when the integrand has a derivative that works nicely when multiplied by xx. We use the same idea here. Let u=arccos⁡yu = \arccos y and dv=dy\dee{v} = \dee{y}, so that v=yv=y and du=−dy1−y2\dee{u} = -\frac{\dee{y}}{\sqrt{1-y^2}}.

∫arccos⁡y dy=yarccos⁡y⏟uv+∫y1−y2dy⏟−vdu\begin{align*}\int\arccos y\,\dee{y} &= \underbrace{y \arccos y}_{uv} +\int \underbrace{\frac{y}{\sqrt{1-y^2}}\dee{y}}_{-v\dee{u}}\end{align*}

Using the substitution u=1−y2u=1-y^2, du=−2ydy\dee{u}=-2y\dee{y},

=yarccos⁡y−12∫u−1/2du=yarccos⁡y−u1/2+C=yarccos⁡y−1−y2+C\begin{align*}&=y\arccos y -\frac{1}{2} \int u^{-1/2}\dee{u}\\ &=y\arccos y - u^{1/2} +C \\&= y \arccos y - \sqrt{1-y^2} + C\end{align*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

Q18Stage 3Past exam · 2016Q3

Evaluate ∫4yarctan⁡(2y) dy\displaystyle\int 4y\arctan(2y) \,\dee{y}.

Hint

After integrating by parts, do some algebraic manipulation to the integral until it's clear how to evaluate it.

Answer

2y2arctan⁡(2y)−y+12arctan⁡(2y)+C2y^2\arctan(2y) - y + \frac12\arctan(2y) + C

Full solution

We integrate by parts, using u=arctan⁡(2y)u = \arctan(2y), dv=4y dy\dee{v} = 4y \,\dee{y}, so that v=2y2v=2y^2 and du=2 dy1+(2y)2\dee{u}= \frac{2\, \dee{y}}{1+(2y)^2}:

∫4yarctan⁡(2y) dy=2y2arctan⁡(2y)⏟uv−∫4y2(2y)2+1 dy⏟vdu\begin{align*} \int 4y\arctan(2y) \,\dee{y} = \underbrace{2y^2\arctan(2y)}_{uv} - \int\underbrace{ \frac{4y^2}{(2y)^2+1} \,\dee{y}}_{v\dee{u}} \end{align*}

The integrand 4y2(2y)2+1\frac{4y^2}{(2y)^2+1} is a rational function. So the remaining integral can be evaluated using the method of partial fractions, starting with long division. But it is easier to just notice that 4y24y2+1=4y2+14y2+1−14y2+1\frac{4y^2}{4y^2+1} = \frac{4y^2+1}{4y^2+1} - \frac{1}{4y^2+1}. We therefore have:

∫4y24y2+1 dy=∫(1−14y2+1) dy=y−12arctan⁡(2y)+C\begin{align*} \int \frac{4y^2}{4y^2+1} \,\dee{y} = \int \left(1 - \frac{1}{4y^2+1}\right)\,\dee{y} = y - \frac12\arctan(2y) + C \end{align*}

The final answer is then

∫4yarctan⁡(2y) dy=2y2arctan⁡(2y)−y+12arctan⁡(2y)+C\begin{align*} \int 4y\arctan(2y) \,\dee{y} = 2y^2\arctan(2y) - y + \frac{1}{2}\arctan(2y) + C \end{align*}
Q19Stage 3

Evaluate ∫x2arctan⁡x dx\displaystyle\int x^2\arctan x ~\dee{x}.

Hint

After integration by parts, use a substitution.

Answer

x33arctan⁡x−16(1+x2)+16log⁡(1+x2)+C\dfrac{x^3}{3}\arctan x- \dfrac{1}{6}(1+x^2) + \dfrac{1}{6}\log(1+x^2)+C

Full solution

We've got an integrand that consists of two functions multiplied together, and no obvious substitution. So, we think about integration by parts. Let's consider our options. Note in Example 1.7.9 of the CLP-2 text, we found that the antiderivative of arctangent is xarctan⁡x−12log⁡(1+x2)+Cx\arctan x -\frac{1}{2}\log(1+x^2)+C.

Option 1:u=arctan⁡xdu=11+x2 dxdv=x2 dxv=13x3Option 2:u=xarctan⁡x−12log⁡(1+x2)du=x2 dxdv=2x dxv=13x3\begin{array}{c|l|l|}\hline \text{Option 1:}& u=\arctan x & \mathrm{d}u=\dfrac{1}{1+x^2}\,\mathrm{d}x\\ & \mathrm{d}v=x^2\,\mathrm{d}x & v=\dfrac{1}{3}x^3\\\hline \text{Option 2:}& u=x\arctan x -\dfrac{1}{2}\log(1+x^2) & \mathrm{d}u=x^2\,\mathrm{d}x\\ & \mathrm{d}v=2x\,\mathrm{d}x & v=\dfrac{1}{3}x^3\\\hline\end{array}

  • Option 1 seems likelier. Let's see how it plays out. We use integration by parts with u=arctan⁡xu=\arctan x, dv=x2dx\dee{v}=x^2 \dee{x}, du=dx1+x2\dee{u}=\frac{\dee{x}}{1+x^2}, and v=13x3v=\frac{1}{3}x^3.

    ∫x2arctan⁡x dx=x33arctan⁡x⏟uv−∫x33(1+x2)dx⏟vdu=x33arctan⁡x−13∫x31+x2dx\begin{align*}\displaystyle\int x^2\arctan x ~\dee{x}&=\underbrace{\frac{x^3}{3}\arctan x}_{uv} - \int \underbrace{\frac{x^3}{3(1+x^2)}\dee{x}}_{v\dee{u}}\\ &=\frac{x^3}{3}\arctan x- \frac{1}{3}\int \frac{x^3}{1+x^2}\dee{x}\end{align*}

    This is starting to look like a candidate for a substitution! Let's try the denominator, s=1+x2s =1+x^2. Then ds=2xdx\dee{s} = 2x\dee{x}, and x2=s−1x^2 = s-1.

    =x33arctan⁡x−16∫x21+x2⋅2xdx=x33arctan⁡x−16∫s−1sds=x33arctan⁡x−16∫1−1s ds=x33arctan⁡x−16s+16log⁡∣s∣+C=x33arctan⁡x−16(1+x2)+16log⁡(1+x2)+C\begin{align*}&= \frac{x^3}{3}\arctan x- \frac{1}{6}\int \frac{x^2}{\textcolor{red}{1+x^2}}\cdot \textcolor{blue}{2x\dee{x}}\\ &=\frac{x^3}{3}\arctan x- \frac{1}{6}\int \frac{s-1}{\textcolor{red}{s}}\textcolor{blue}{\dee{s}}\\ &=\frac{x^3}{3}\arctan x- \frac{1}{6} \int 1 - \frac{1}{s}~\dee{s}\\ &=\frac{x^3}{3}\arctan x- \frac{1}{6}s + \frac{1}{6}\log|s|+C\\ &=\frac{x^3}{3}\arctan x- \frac{1}{6}(1+x^2) + \frac{1}{6}\log(1+x^2)+C\end{align*}

  • What if we had tried the other option? That is, u=x2u=x^2, du=2xdx\dee{u} = 2x\dee{x}, dv=arctan⁡x\dee{v} = \arctan x, and v=xarctan⁡x−12log⁡(1+x2)v = x\arctan x - \frac{1}{2}\log(1+x^2). It's not always the case that both options work, but sometimes they do. (They are almost never of equal difficulty.) This solution takes advantage of two previously hard-won results: the antiderivatives of logarithm and arctangent.

∫x2arctan⁡xdx=x2⏟u(xarctan⁡x−12log⁡(1+x2))⏟v−∫(xarctan⁡x−12log⁡(1+x2))⏟v⋅2xdx⏟du=x3arctan⁡x−x22log⁡(1+x2)−2∫x2arctan⁡xdx+∫xlog⁡(1+x2)dx\begin{align*}\int x^2\arctan x \dee{x}&=\underbrace{x^2}_{u}\underbrace{\left(x\arctan x - \frac{1}{2}\log(1+x^2)\right)}_v - \int \underbrace{\left(x\arctan x - \frac{1}{2}\log(1+x^2)\right)}_v\cdot \underbrace{2x\dee{x}}_{\dee{u}}\\ &=x^3\arctan x - \frac{x^2}{2}\log(1+x^2)- 2\int x^2\arctan x \dee{x} + \int x\log(1+x^2)\dee{x}\end{align*}

Adding 2∫x2arctan⁡xdx2\displaystyle\int x^2\arctan x \dee{x} to both sides:

3∫x2arctan⁡xdx=x3arctan⁡x−x22log⁡(1+x2)+∫xlog⁡(1+x2)dx∫x2arctan⁡xdx=x33arctan⁡x−x26log⁡(1+x2)+13∫xlog⁡(1+x2)dx\begin{align*}3\int x^2\arctan x \dee{x}&=x^3\arctan x - \frac{x^2}{2}\log(1+x^2) + \int x\log(1+x^2)\dee{x}\\ \int x^2\arctan x \dee{x}&=\frac{x^3}{3}\arctan x - \frac{x^2}{6}\log(1+x^2) + \frac{1}{3}\int x\log(1+x^2)\dee{x}\end{align*}

Using the substitution s=1+x2s=1+x^2, ds=2xdx\dee{s}=2x\dee{x}:

=x33arctan⁡x−x26log⁡(1+x2)+16∫log⁡sds\begin{align*}&=\frac{x^3}{3}\arctan x - \frac{x^2}{6}\log(1+x^2) + \frac{1}{6}\int\log s\dee{s}\end{align*}

Using the antiderivative of logarithm found in Example 1.7.8 of the CLP-2 text,

=x33arctan⁡x−x26log⁡(1+x2)+16(slog⁡s−s)+C=x33arctan⁡x−x26log⁡(1+x2)+16((1+x2)log⁡(1+x2)−(1+x2))+C=x33arctan⁡x+[−x26+1+x26]log⁡(1+x2)−16(1+x2)+C=x33arctan⁡x+16log⁡(1+x2)−16(1+x2)+C\begin{align*}&=\frac{x^3}{3}\arctan x - \frac{x^2}{6}\log(1+x^2) + \frac{1}{6}\left(s\log s - s\right)+C\\ &=\frac{x^3}{3}\arctan x - \frac{x^2}{6}\log(1+x^2) + \frac{1}{6}\left((1+x^2)\log (1+x^2) - (1+x^2)\right)+C\\ &=\frac{x^3}{3}\arctan x + \left[-\frac{x^2}{6}+\frac{1+x^2}{6}\right]\log(1+x^2)-\frac{1}{6}(1+x^2)+C\\ &=\frac{x^3}{3}\arctan x +\frac{1}{6}\log(1+x^2)-\frac{1}{6}(1+x^2)+C\end{align*}
Q20Stage 3

Evaluate ∫ex/2cos⁡(2x)dx\displaystyle\int e^{x/2}\cos(2x)\dee{x}.

Hint

This example is similar to Example 1.7.10 in the CLP-2 text. The functions ex/2e^{x/2} and cos⁡(2x)\cos(2x) both do not substantially alter when we differentiate or antidifferentiate them. If we use integration by parts twice, we'll end up with an expression that includes our original integral. Then we can just solve for the original integral in the equation, without actually integrating.

Answer

217ex/2cos⁡(2x)+817ex/2sin⁡(2x)+C\dfrac{2}{17}e^{x/2}\cos(2x)+\dfrac{8}{17}e^{x/2}\sin(2x)+C

Full solution

This example is similar to Example 1.7.10 in the CLP-2 text. The functions ex/2e^{x/2} and cos⁡(2x)\cos(2x) both do not substantially alter when we differentiate or antidifferentiate them. If we use integration by parts twice, we'll end up with an expression that includes our original integral. Then we can just solve for the original integral in the equation, without actually antidifferentiating.

Let's use u=cos⁡(2x)u=\cos(2x) and dv=ex/2dx\dee{v} = e^{x/2}\dee{x}, so du=−2sin⁡(2x)dx\dee{u}=-2\sin(2x)\dee{x} and v=2ex/2v=2e^{x/2}.

∫ex/2cos⁡(2x)dx=2ex/2cos⁡(2x)⏟uv−∫−4ex/2sin⁡(2x)dx⏟vdu=2ex/2cos⁡(2x)+4∫ex/2sin⁡(2x)dx\begin{align*}\int e^{x/2}\cos(2x)\dee{x}&= \underbrace{2e^{x/2}\cos(2x)}_{uv}- \int\underbrace{-4e^{x/2}\sin(2x)\dee{x}}_{v\dee{u}}\\ &=2e^{x/2}\cos(2x)+4\int e^{x/2}\sin(2x)\dee{x}\end{align*}

Similarly to our first integration by parts, we use u=sin⁡(2x)u=\sin(2x), dv=ex/2dx\dee{v}=e^{x/2}\dee{x}, du=2cos⁡(2x)dx\dee{u}=2\cos(2x)\dee{x}, and v=2ex/2v=2e^{x/2}.

=2ex/2cos⁡(2x)+4[2ex/2sin⁡(2x)⏟uv−∫4ex/2cos⁡(2x)dx⏟vdu]\begin{align*}&=2e^{x/2}\cos(2x)+4\left[ \underbrace{2e^{x/2}\sin(2x)}_{uv} -\int\underbrace{4e^{x/2}\cos(2x)\dee{x}}_{v\dee{u}} \right]\end{align*}

So, we've found the equation

∫ex/2cos⁡(2x)dx=2ex/2cos⁡(2x)+8ex/2sin⁡(2x)−16∫ex/2cos⁡(2x)dx+C\begin{align*}\color{red}\int e^{x/2}\cos(2x)\dee{x}&=2e^{x/2}\cos(2x)+8e^{x/2}\sin(2x)-\color{red}16\int e^{x/2}\cos(2x)\dee{x}\color{black}+C\end{align*}

We add 16∫ex/2cos⁡(2x)dx16\displaystyle\int e^{x/2}\cos(2x)\dee{x} to both sides.

17∫ex/2cos⁡(2x)dx=2ex/2cos⁡(2x)+8ex/2sin⁡(2x)+C∫ex/2cos⁡(2x)dx=217ex/2cos⁡(2x)+817ex/2sin⁡(2x)+C\begin{align*}\color{red}17\int e^{x/2}\cos(2x)\dee{x}&=2e^{x/2}\cos(2x)+8e^{x/2}\sin(2x)+C \\ \int e^{x/2}\cos(2x)\dee{x}&=\frac{2}{17}e^{x/2}\cos(2x)+\frac{8}{17}e^{x/2}\sin(2x)+C\end{align*}

Remark: remember that CC is a stand-in for “we can add any real constant". Since CC can be any number in (−∞,∞)(-\infty,\infty), also C17\frac{C}{17} can be any number in (−∞,∞)(-\infty,\infty). So, rather than write C17\frac{C}{17} in the last line, we re-named C17\frac{C}{17} to CC.

Q21Stage 3

Evaluate ∫sin⁡(log⁡x)dx\displaystyle\int \sin(\log x)\dee{x}.

Hint

This looks a bit like a substitution problem, because we have an “inside function."

It might help to review Example 1.7.11 in the CLP-2 text.

Answer

x2[sin⁡(log⁡x)−cos⁡(log⁡x)]+C\dfrac{x}{2} \big[\sin(\log x) - \cos (\log x)\big]+C

Full solution

  • This question looks like a substitution, since we have an “inside function." So, let's see where that leads: let u=log⁡xu=\log x. Then du=1x dx\dee{u}=\dfrac{1}{x}~\dee{x}. We don't see this right away in our function, but we can bring it into the function by multiplying and dividing by xx, and noting from our substitution that eu=xe^u=x.

    ∫sin⁡(log⁡x)dx=∫xsin⁡(log⁡x)xdx=∫eusin⁡u du\begin{align*}\int \sin(\log x)\dee{x}&=\int \frac{x\sin(\log x)}{x}\dee{x}\\ &=\int e^u \sin u ~\dee{u}\end{align*}

    Using the result of Example 1.7.11 in the CLP-2 text:

    =12eu(sin⁡u−cos⁡u)+C=12elog⁡x(sin⁡(log⁡x)−cos⁡(log⁡x))+C=12x(sin⁡(log⁡x)−cos⁡(log⁡x))+C\begin{align*}&=\frac{1}{2}e^u\left(\sin u - \cos u\right)+C\\ &=\frac{1}{2}e^{\log x}\left(\sin(\log x) - \cos (\log x)\right)+C \\&=\frac{1}{2}x \left(\sin(\log x) - \cos (\log x)\right)+C\end{align*}

  • It's not clear how to antidifferentiate the integrand, but we can certainly differentiate it. So, keeping in mind the method of Example 1.7.11 in the CLP-2 text, we take u=sin⁡(log⁡x)u=\sin(\log x) and dv=dx\dee{v}=\dee{x}, so du=1xcos⁡(log⁡x)dx\dee{u}=\frac{1}{x}\cos(\log x)\dee{x} and v=xv=x.

    ∫sin⁡(log⁡x)dx=xsin⁡(log⁡x)⏟uv−∫cos⁡(log⁡x)dx⏟vdu\begin{align*}\int \sin(\log x)\dee{x}&= \underbrace{x\sin(\log x)}_{uv} - \int \underbrace{ \cos(\log x)\dee{x} }_{v\dee{u}}\end{align*}

    Continuing on, we again use integration by parts, with u=cos⁡(log⁡x)u=\cos(\log x), dv=dx\dee{v}=\dee{x}, du=−1xsin⁡(log⁡x)dx\dee{u}=-\frac{1}{x}\sin(\log x)\dee{x}, and v=xv=x.

    =xsin⁡(log⁡x)−[xcos⁡(log⁡x)⏟uv+∫sin⁡(log⁡x)⏟−vdudx]\begin{align*}&=x\sin(\log x) - \bigg[\underbrace{x\cos(\log x)}_{uv} + \int\underbrace{\sin(\log x)}_{-v\dee{u}}\dee{x} \bigg]\end{align*}

    That is, we have

    ∫sin⁡(log⁡x)dx=x[sin⁡(log⁡x)−cos⁡(log⁡x)]−∫sin⁡(log⁡x)dx+C\begin{align*}\int \sin(\log x)\dee{x}&=x\left[\sin(\log x) - \cos(\log x)\right] - \int \sin(\log x)\dee{x}+C\end{align*}

    Adding ∫sin⁡(log⁡x)dx\int \sin(\log x)\dee{x} to both sides,

    2∫sin⁡(log⁡x)dx=x[sin⁡(log⁡x)−cos⁡(log⁡x)]+C∫sin⁡(log⁡x)dx=x2[sin⁡(log⁡x)−cos⁡(log⁡x)]+C\begin{align*}2\int \sin(\log x)\dee{x}&=x\left[\sin(\log x) - \cos(\log x)\right]+C \\\int \sin(\log x)\dee{x}&=\frac{x}{2}\left[\sin(\log x) - \cos(\log x)\right]+C\end{align*}

    Remark: remember that CC is a stand-in for “we can add any real constant". Since CC can be any number in (−∞,∞)(-\infty,\infty), also C2\frac{C}{2} can be any number in (−∞,∞)(-\infty,\infty). So, rather than write C2\frac{C}{2} in the last line, we re-named C2\frac{C}{2} to CC.

Q22Stage 3

Evaluate ∫2x+log⁡2xdx\displaystyle\int 2^{x+\log_2 x} \dee{x}.

Hint

Start by simplifying.

Answer

2xlog⁡2(x−1log⁡2)+C\dfrac{2^x}{\log 2}\left(x - \dfrac{1}{\log 2}\right)+C

Full solution

We begin by simplifying the integrand.

∫2x+log⁡2xdx=∫2x⋅2log⁡2xdx=∫2x⋅x dx\begin{align*}\int 2^{x+\log_2 x} \dee{x}&= \int 2^{x}\cdot 2^{\log_2 x} \dee{x}= \int 2^{x}\cdot x~ \dee{x}\end{align*}

This is similar to the integral ∫xexdx\displaystyle\int xe^x \dee{x}, which we saw in Example 1.7.1 of the CLP-2 text. Let's write 2=elog⁡22=e^{\log 2} to take advantage of the easy integrability of exe^x.

=∫x⋅exlog⁡2dx\begin{align*}&=\int x\cdot e^{x\log 2} \dee{x}\end{align*}

We use integration by parts with u=xu=x, dv=exlog⁡2dx\dee{v}=e^{x\log 2}\dee{x}; du=dx\dee{u}=\dee{x}, v=1log⁡2exlog⁡2v = \frac{1}{\log 2}e^{x\log 2}. (Remember log⁡2\log 2 is a constant. If you'd prefer, you can do a substitution with s=xlog⁡2s= x\log 2 first, to have a simpler exponent of ee.)

=xlog⁡2exlog⁡2⏟uv−∫1log⁡2exlog⁡2dx⏟vdu=xlog⁡2exlog⁡2−1(log⁡2)2exlog⁡2+C=xlog⁡22x−1(log⁡2)22x+C\begin{align*}&=\underbrace{\frac{x}{\log 2}e^{x\log 2}}_{uv} - \int \underbrace{\frac{1}{\log 2}e^{x\log 2}\dee{x}}_{v\dee{u}}\\ &=\frac{x}{\log 2}e^{x\log 2} - \frac{1}{(\log 2)^2}e^{x\log 2}+C\\ &=\frac{x}{\log 2}2^x - \frac{1}{(\log 2)^2}2^x+C\end{align*}
Q23Stage 3

Evaluate ∫ecos⁡xsin⁡(2x)dx\displaystyle\int e^{\cos x}\sin(2x)\dee{x}.

Hint

sin⁡(2x)=2sin⁡xcos⁡x\sin(2x) = 2\sin x \cos x

Answer

2ecos⁡x[1−cos⁡x]+C2e^{\cos x}[1-\cos x]+C

Full solution

It's not obvious where to start, but in general it's nice to have the arguments of our trig functions the same. So, we use the identity sin⁡(2x)=2sin⁡xcos⁡x\sin(2x)=2\sin x \cos x.

∫ecos⁡xsin⁡(2x)dx=2∫ecos⁡xcos⁡xsin⁡x dx\begin{align*}\int e^{\cos x}\sin(2x)\dee{x}&=2\int e^{\cos x}\cos x \sin x\,\dee{x}\end{align*}

Now we can use the substitution w=cos⁡xw=\cos x, dw=−sin⁡xdx\dee{w}=-\sin x \dee{x}.

=−2∫wewdw\begin{align*}&=-2\int we^w\dee{w}\end{align*}

From here the integral should look more familiar. We can use integration by parts with u=wu=w, dv=ewdw\dee{v} = e^w\dee{w}, du=dw\dee{u}=\dee{w}, and v=ewv=e^w.

=−2[wew⏟uv−∫ewdw⏟vdu]=2ew[1−w]+C=2ecos⁡x[1−cos⁡x]+C\begin{align*}&=-2\left[\underbrace{we^w}_{uv} - \int \underbrace{e^w\dee{w}}_{v\dee{u}}\right]\\ &=2e^w\left[1-w\right]+C\\ &=2e^{\cos x}[1-\cos x]+C\end{align*}
Q24Stage 3

Evaluate ∫xe−x(1−x)2dx\displaystyle\int \frac{x e^{-x}}{(1-x)^2}\dee{x}.

Hint

What is the derivative of xe−xx e^{-x}?

Answer

∫xe−x(1−x)2dx=xe−x1−x+e−x+C=e−x1−x+C\displaystyle\int \frac{x e^{-x}}{(1-x)^2}\dee{x} = \frac{xe^{-x}}{1-x} + e^{-x} + C =\frac{e^{-x}}{1-x} + C

Full solution

We've got an integrand that consists of several functions multiplied together, and no obvious substitution. So, we think about integration by parts. We know an antiderivative for 1(1−x)2\frac{1}{(1-x)^2}, because we know ddx11−x=1(1−x)2\diff{}{x}\frac{1}{1-x}=\frac{1}{(1-x)^2}. So let's try dv=dx(1−x)2\dee{v} = \frac{\dee{x}}{(1-x)^2} and u=xe−xu=xe^{-x}. Then v=11−xv=\frac{1}{1-x} and du=(1−x)e−x dx\dee{u} = (1-x)e^{-x}\,\dee{x}. So, by integration by parts,

∫xe−x⏟udx(1−x)2⏟dv=xe−x1−x⏟uv−∫11−x⏟v(1−x)e−x dx⏟du=xe−x1−x−∫e−x dx=xe−x1−x+e−x+C=e−x1−x+C\begin{align*} \int\underbrace{xe^{-x}}_{u}\underbrace{ \frac{\dee{x}}{(1-x)^2}}_{\dee{v}} &= \underbrace{\frac{xe^{-x}}{1-x}}_{uv} -\int\underbrace{\frac{1}{1-x}}_{v}\underbrace{(1-x)e^{-x}\,\dee{x}}_{\dee{u}} \\ &=\frac{xe^{-x}}{1-x} -\int e^{-x}\,\dee{x} \\ &=\frac{xe^{-x}}{1-x} + e^{-x} + C =\frac{e^{-x}}{1-x} + C \end{align*}
Q25Stage 3Past exam · 2000D

A reduction formula.

  1. Derive the reduction formula

    ∫sin⁡n(x) dx=−sin⁡n−1(x)cos⁡(x)n+n−1n∫sin⁡n−2(x) dx.\int\sin^n(x)\,\dee{x}=-\frac{\sin^{n-1}(x)\cos(x)}{n} +\frac{n-1}{n}\int\sin^{n-2}(x)\,\dee{x}.

  2. Calculate ∫0π/2sin⁡8(x) dx\displaystyle\int_0^{\pi/2}\sin^8(x)\,\dee{x}.

Hint

You'll want to do an integration by parts for (a)–check the end result to get a guess as to what your parts should be. A trig identity and some amount of algebraic manipulation will be necessary to get the final form.

Answer

(a) We integrate by parts with u=sin⁡n−1xu=\sin^{n-1}x and dv=sin⁡x dx\dee{v}=\sin x\,\dee{x}, so that du=(n−1)sin⁡n−2xcos⁡x\dee{u}=(n-1)\sin^{n-2}x\cos x and v=−cos⁡xv=-\cos x.

∫sin⁡nx dx=−sin⁡n−1x cos⁡x⏟uv+(n−1)∫cos⁡2x sin⁡n−2x dx⏟−∫vdu\begin{align*}\int\sin^nx\,\dee{x} &=\underbrace{-\sin^{n-1}x\ \cos x}_{uv}+\underbrace{(n-1)\int \cos^2x\ \sin^{n-2}x\ \dee{x}}_{-\int v\dee{u}}\end{align*}

Using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x =1,

=−sin⁡n−1x cos⁡x+(n−1)∫(1−sin⁡2x)sin⁡n−2x dx=−sin⁡n−1x cos⁡x+(n−1)∫sin⁡n−2x dx−(n−1)∫sin⁡nx dx\begin{align*}&=-\sin^{n-1}x\ \cos x+(n-1)\int (1-\sin^2 x)\sin^{n-2}x\ \dee{x}\\ &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \dee{x} -(n-1)\int\sin^{n}x\ \dee{x}\end{align*}

Moving the last term on the right hand side to the left hand side gives

n∫sin⁡nx dx=−sin⁡n−1x cos⁡x+(n−1)∫sin⁡n−2x dx\begin{align*} n\int\sin^nx\,\dee{x} &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \dee{x} \end{align*}

Dividing across by nn gives the desired reduction formula.

(b) 35256π≈0.4295\dfrac{35}{256}\pi\approx0.4295

Full solution

(a) The “parts" in the integrand are powers of sine. Looking at the right hand side of the reduction formula, we see that it looks a little like the derivative of sin⁡n−1x\sin^{n-1}x, although not exactly. So, let's integrate by parts with u=sin⁡n−1xu=\sin^{n-1}x and dv=sin⁡x dx\dee{v}=\sin x\,\dee{x}, so that du=(n−1)sin⁡n−2xcos⁡x\dee{u}=(n-1)\sin^{n-2}x\cos x and v=−cos⁡xv=-\cos x.

∫sin⁡nx dx=−sin⁡n−1x cos⁡x⏟uv+(n−1)∫cos⁡2x sin⁡n−2x dx⏟−∫vdu\begin{align*}\int\sin^nx\,\dee{x} &=\underbrace{-\sin^{n-1}x\ \cos x}_{uv}+\underbrace{(n-1)\int \cos^2x\ \sin^{n-2}x\ \dee{x}}_{-\int v\dee{u}}\end{align*}

Using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x =1,

=−sin⁡n−1x cos⁡x+(n−1)∫(1−sin⁡2x)sin⁡n−2x dx=−sin⁡n−1x cos⁡x+(n−1)∫sin⁡n−2x dx−(n−1)∫sin⁡nx dx\begin{align*}&=-\sin^{n-1}x\ \cos x+(n-1)\int (1-\sin^2 x)\sin^{n-2}x\ \dee{x}\\ &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \dee{x} -(n-1)\int\sin^{n}x\ \dee{x}\end{align*}

Moving the last term on the right hand side to the left hand side gives

n∫sin⁡nx dx=−sin⁡n−1x cos⁡x+(n−1)∫sin⁡n−2x dx\begin{align*} n\int\sin^nx\,\dee{x} &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \dee{x} \end{align*}

Dividing across by nn gives the desired reduction formula.

(b) By the reduction formula of part (a), if n≥2n \ge 2,

∫0π/2sin⁡n(x) dx=n−1n∫0π/2sin⁡n−2(x) dx\begin{align*} \int_0^{\pi/2}\sin^n(x)\,\dee{x}= \frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}(x)\,\dee{x} \end{align*}

since sin⁡0=cos⁡π2=0\sin 0=\cos\frac{\pi}{2}=0. Applying this reduction formula, with n=8,6,4,2n=8,6,4,2:

∫0π/2sin⁡8(x) dx=78∫0π/2sin⁡6(x) dx=78⋅56∫0π/2sin⁡4(x) dx=78⋅56⋅34∫0π/2sin⁡2(x) dx=78⋅56⋅34⋅12∫0π/2 dx=78⋅56⋅34⋅12⋅π2=35256π\begin{align*} \int_0^{\pi/2}\sin^8(x)\,\dee{x} &=\frac{7}{8}\int_0^{\pi/2}\sin^6(x)\,\dee{x} =\frac{7}{8}\cdot\frac{5}{6}\int_0^{\pi/2}\sin^4(x)\,\dee{x} =\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\int_0^{\pi/2}\sin^2(x)\,\dee{x} \\ &=\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\int_0^{\pi/2}\,\dee{x} =\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} =\frac{35}{256}\pi \end{align*}

Using a calculator, we see this is approximately 0.42950.4295.

Q26Stage 3Past exam · 1996A

Let RR be the part of the first quadrant that lies below the curve y=arctan⁡xy=\arctan x and between the lines x=0x=0 and x=1x=1.

  1. Sketch the region RR and determine its area.

  2. Find the volume of the solid obtained by rotating RR about the yy–axis.

Hint

See Examples 1.7.9 and 1.6.5 in the

CLP-2 text for refreshers on integrating arctangent, and using washers.

Remember tan⁡2x+1=sec⁡2x\tan^2 x + 1 = \sec^2 x, and sec⁡2x\sec^2x is easy to integrate.

Answer

(a) Area: π4−log⁡22\dfrac{\pi}{4}-\dfrac{\log 2}{2}

Figure from prob_s1.7, line 886

Figure from prob_s1.7, line 886

(b) Volume: π22−π\dfrac{\pi^2}{2}-\pi

Full solution

(a) The sketch is the figure on the left below. By integration by parts with u=arctan⁡xu=\arctan x, dv=dx\dee{v}=\dee{x}, v=xv=x and du=11+x2 dx\dee{u}=\frac{1}{1+x^2}\,\dee{x}, and then the substitution s=1+x2s=1+x^2,

A=∫01arctan⁡x dx=xarctan⁡x⏟uv∣01−∫01x1+x2 dx⏟vdu=arctan⁡1−12log⁡(1+x2)∣01=π4−log⁡22\begin{align*} A&=\int_0^1\arctan x\ \dee{x}=\underbrace{ x\arctan x}_{uv}\Big|_0^1-\int_0^1\underbrace{\frac{x}{1+x^2}\,\dee{x}}_{v\dee{u}} =\arctan 1-\half\log(1+x^2)\Big|_0^1\\ &=\frac{\pi}{4}-\frac{\log 2}{2} \end{align*}

Figure from prob_s1.7, line 886

Figure from prob_s1.7, line 886

Figure from prob_s1.7, line 893

Figure from prob_s1.7, line 893

(b) We'll use horizontal washers as in Example 1.6.5 of the CLP-2 text.

  • We cut RR into thin horizontal strips of width dy\dee{y} as in the figure on the right above.

  • When we rotate RR about the yy–axis, each strip sweeps out a thin washer

    • whose inner radius is rin=tan⁡yr_{in}=\tan y and outer radius is rout=1r_{out}=1, and

    • whose thickness is dy\dee{y} and hence

    • whose volume π(rout2−rin2)dy=π(1−tan⁡2y)dy\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi(1-\tan^2 y)\dee{y}.

  • As our bottommost strip is at y=0y=0 and our topmost strip is at y=π4y=\frac{\pi}{4} (since at the top x=1x=1 and x=tan⁡yx=\tan y), the total

    Volume=∫0π ⁣/4π(1−tan⁡2y) dy=∫0π ⁣/4π(2−sec⁡2y) dy=π[2y−tan⁡y]0π ⁣/4=π22−π\begin{align*} \text{Volume} &= \int _0^{\nicefrac{\pi}{4}} \pi(1-\tan^2 y)\ \dee{y} = \int _0^{\nicefrac{\pi}{4}} \pi(2-\sec^2 y)\ \dee{y} =\pi\big[2y -\tan y\big]_0^{\nicefrac{\pi}{4}} \\ &= \frac{\pi^2}{2}-\pi \end{align*}
Q27Stage 3Past exam · 2016Q3

Let RR be the region between the curves T(x)=xe3xT(x) = \sqrt{x}e^{3x} and B(x)=x(1+2x)B(x) = \sqrt{x}(1+2x) on the interval 0≤x≤30 \le x \le 3. (It is true that T(x)≥B(x)T(x)\ge B(x) for all 0≤x≤30\le x\le 3.) Compute the volume of the solid formed by rotating RR about the xx-axis.

Hint

Your integral can be broken into two integrals, which yield to two different integration methods.

Answer

π(17e18−437336)\pi \left( \dfrac{17 e^{18}-4373}{36} \right)

Full solution

For a fixed value of xx, if we rotate about the xx-axis, we form a washer of inner radius B(x)B(x) and outer radius T(x)T(x) and hence of area π[T(x)2−B(x)2]\pi [T(x)^2 - B(x)^2]. We integrate this function from x=0x=0 to x=3x=3 to find the total volume VV:

V=∫03π[T(x)2−B(x)2] dx=π∫03(xe3x)2−(x(1+2x))2 dx=π∫03(xe6x−(x+4x2+4x3)) dx=π∫03xe6x dx−π[x22+4x33+x4]03=π∫03xe6x dx−π[322+4⋅333+34]\begin{align*} V &= \int_0^3 \pi [T(x)^2 - B(x)^2]\,\dee{x} \\ &= \pi \int_0^3 (\sqrt{x}e^{3x})^2 - (\sqrt{x}(1+2x))^2 \,\dee{x} \\ &= \pi \int_0^3 \big( xe^{6x} - (x+4x^2+4x^3) \big) \,\dee{x} \\ &= \pi \int_0^3 xe^{6x} \,\dee{x} - \pi \Big[ \frac{x^2}{2} + \frac{4x^3}{3} + x^4\Big]_{0}^3 \\ &= \pi \int_0^3 xe^{6x} \,\dee{x} - \pi \Big[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4\Big] \end{align*}

For the first integral, we use integration by parts with u(x)=xu(x) = x, dv=e6xdx\dee{v} = e^{6x}\dee{x}, so that du=dx\dee{u}=\dee{x} and v(x)=16e6xv(x)=\frac16e^{6x}:

∫03xe6x dx=xe6x6⏟uv∣03−∫0316e6x dx⏟vdu=3e186−0−136e6x∣03=e182−(e1836−136).\begin{align*} \int_0^3 xe^{6x} \,\dee{x} &=\underbrace{ \frac{xe^{6x}}{6}}_{uv}\bigg|_{0}^3 - \int_0^3 \underbrace{\frac{1}{6}e^{6x} \,\dee{x} }_{v\dee{u}}\\ &= \frac{3e^{18}}{6} - 0 - \frac{1}{36} e^{6x} \bigg|_{0}^3 = \frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg). \end{align*}

Therefore, the total volume is

V=π[e182−(e1836−136)]−π[322+4⋅333+34]=π(17e18−437336).\begin{align*} V = \pi \bigg[\frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg) \bigg] - \pi \bigg[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4 \bigg] = \pi \bigg( \frac{17 e^{18}-4373}{36} \bigg). \end{align*}
Q28Stage 3Past exam · M105 2013A

Let f(0)=1f(0) = 1, f(2)=3f(2) = 3 and f′(2)=4f'(2) = 4. Calculate ∫04f′′(x) dx{\displaystyle\int_0^4 f''\big(\sqrt{x}\big)\,\dee{x}}.

Hint

Think, first, about how to get rid of the square root in the argument of f′′f'', and, second, how to convert f′′f'' into f′f'. Note that you are told that f′(2)=4f'(2) = 4 and f(0)=1f(0) = 1, f(2)=3f(2) = 3.

Answer

1212

Full solution

To get rid of the square root in the argument of f′′f'', we make the change of variables (also called “substitution") x=t2, dx=2t dtx=t^2,\ \dee{x}=2t\,\dee{t}.

∫04f′′(x) dx=2∫02tf′′(t) dt\begin{align*} \int_0^4 f''\big(\sqrt{x}\big)\,\dee{x} &= 2\int_0^2 tf''(t)\,\dee{t} \end{align*}

Then, to convert f′′f'' into f′f', we integrate by parts with u=tu=t, dv=f′′(t) dt\dee{v}=f''(t)\,\dee{t}, v=f′(t)v=f'(t).

∫04f′′(x) dx=2{[tf′(t)⏟uv]02−∫02 ⁣ ⁣ ⁣f′(t) dt⏟vdu}=2[tf′(t)−f(t)]02=2[2f′(2)−f(2)+f(0)]=2[2×4−3+1]=12\begin{align*} \int_0^4 f''\big(\sqrt{x}\big)\,\dee{x} &= 2\bigg\{\Big[\underbrace{tf'(t)}_{uv}\Big]_0^2-\int_0^2\!\!\!\underbrace{ f'(t)\,\dee{t}}_{v\dee{u}}\bigg\} \\ &=2\Big[tf'(t)-f(t)\Big]_0^2 \\ &=2\big[2f'(2)-f(2)+f(0)\big]=2\big[2\times 4-3+1\big]\\ &=12 \end{align*}
Q29Stage 3

Evaluate lim⁡n→∞∑i=1n2n(2ni−1)e2ni−1\displaystyle\lim_{n \to \infty}\sum_{i=1}^n \frac{2}{n}\left(\frac{2}{n}i-1\right)e^{\frac{2}{n}i-1} .

Hint

Interpret the limit as a right Riemann sum.

Answer

2e\dfrac{2}{e}

Full solution

As we saw in Section 1.1 of the CLP-2 text, there are many different ways to interpret a limit as a Riemann sum. In the absence of instructions that restrain our choices, we go with the most convenient interpretations.

With that in mind, we choose:

  • that our Riemann sum is a right Riemann sum (because we see ii, not i−1i-1 or i−12i-\frac{1}{2})

  • Δx=2n\Delta x = \frac{2}{n} (because it is multiplied by the rest of the integrand, and also shows up multiplied by ii),

  • then xi=a+iΔx=2ni−1x_i = a+i\Delta x = \frac{2}{n}i-1, which leads us to a=−1a=-1 and

  • f(x)=xexf(x) = xe^x.

  • Finally, since Δx=b−an=2n\Delta x = \frac{b-a}{n}=\frac{2}{n} and a=−1a=-1, we have b=1b=1.

So, the limit is equal to the definite integral

lim⁡n→∞∑i=1n2n(2ni−1)e2ni−1=∫−11xex dx\begin{align*}\lim_{n \to \infty}\sum_{i=1}^n \frac{2}{n}\left(\frac{2}{n}i-1\right)e^{\frac{2}{n}i-1}&=\int_{-1}^1x e^x~\dee{x}\end{align*}

which we evaluate using integration by parts with u=xu=x, dv=exdx\dee{v}=e^x\dee{x}, du=dx\dee{u}=\dee{x}, and v=exv=e^x.

=[xex⏟uv]−11−∫−11exdx⏟vdu=(e+1e)−(e−1e)=2e\begin{align*}&=\Big[\underbrace{xe^x}_{uv} \Big]_{-1}^1 - \int_{-1}^1 \underbrace{e^x\dee{x}}_{v\dee{u}}\\ &=\left(e+\frac{1}{e}\right) - \left(e-\frac{1}{e}\right) = \frac{2}{e}\end{align*}

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.