1.7 Integration by parts

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.

chain; product

Remember our rule: $\int u \dee{v} = uv - \int v \dee{u}$. So, we take $u$ and use it to make $\dee{u}$, and we take $\dee{v}$ and use it to make $v$.

The part chosen as $u$ will be differentiated. The part chosen as $\dee{v}$ will be antidifferentiated.

According to the quotient rule,

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

$\displaystyle\int \frac{f'(x)}{g(x)}~\dee{x}= \dfrac{f(x)}{g(x)} +\displaystyle\int\frac{f(x)g'(x)}{g^2(x)}\dee{x}$

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

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

where $D$ is any constant.

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

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

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

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

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 $\log x$, or differentiate it?

$\dfrac{x^2\log x}{2} - \dfrac{x^2}{4} + C$

This problem is similar to Question 6.

$- \dfrac{\log x}{6 x^6} - \dfrac{1}{36 x^6} + C$

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.

$\pi$

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

$\dfrac{\pi}{2} -1$

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

$e^x\left(x^3-3x^2+6x - 6\right)+C$

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

$\dfrac{x^2}{2}\log^3x - \dfrac{3x^2}{4}\log^2 x + \dfrac{3x^2}{4}\log x - \dfrac{3x^2}{8}+C$

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

$(2-x^2)\cos x + 2x\sin x +C$

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

$\left( t^3 - \frac{5}{2}t^2+6t \right)\log t -\frac{1}{3}t^3 +\frac{5}{4}t^2-6t+C$

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

$e^{\sqrt{s}}\left(2s - 4\sqrt{s} +4\right)+C$

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

$x\log^2 x -2x\log x +2x+C$

Not every integral should be evaluated using integration by parts.

$e^{x^2+1}+C$

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.

$y \arccos y - \sqrt{1-y^2} + C$

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

$2y^2\arctan(2y) - y + \frac12\arctan(2y) + C$

After integration by parts, use a substitution.

$\dfrac{x^3}{3}\arctan x- \dfrac{1}{6}(1+x^2) + \dfrac{1}{6}\log(1+x^2)+C$

This example is similar to Example 1.7.10 in the CLP-2 text. The functions $e^{x/2}$ and $\cos(2x)$ 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.

$\dfrac{2}{17}e^{x/2}\cos(2x)+\dfrac{8}{17}e^{x/2}\sin(2x)+C$

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.

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

Start by simplifying.

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

$\sin(2x) = 2\sin x \cos x$

$2e^{\cos x}[1-\cos x]+C$

What is the derivative of $x e^{-x}$?

$\displaystyle\int \frac{x e^{-x}}{(1-x)^2}\dee{x} = \frac{xe^{-x}}{1-x} + e^{-x} + C =\frac{e^{-x}}{1-x} + C$

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.

(a) We integrate by parts with $u=\sin^{n-1}x$ and $\dee{v}=\sin x\,\dee{x}$, so that $\dee{u}=(n-1)\sin^{n-2}x\cos x$ and $v=-\cos x$.

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

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

Dividing across by $n$ gives the desired reduction formula.

(b) $\dfrac{35}{256}\pi\approx0.4295$

See Examples 1.7.9 and 1.6.5 in the

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

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

(a) Area: $\dfrac{\pi}{4}-\dfrac{\log 2}{2}$

(b) Volume: $\dfrac{\pi^2}{2}-\pi$

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

$\pi \left( \dfrac{17 e^{18}-4373}{36} \right)$

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

$12$

Interpret the limit as a right Riemann sum.

$\dfrac{2}{e}$