1.4 Substitution

One is true, the other false.

(a) true (b) false

(a) This is true: it is an application of Theorem ? in the CLP-2 text with $f(x)=\sin x$ and $u(x)=e^x$.

(b) This is false: the upper limit of integration is incorrect. Using Theorem 1.4.6 in the CLP-2 text, the correct form is

Alternately, we can use the Fundamental Theorem of Calculus Part 2, and our answer from (a):

You can check whether the final answer is correct by differentiating.

The reasoning is not sound: when we do a substitution, we need to take care of the differential ($\dee{x}$). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here's the way to do it:

Problem: Evaluate $\displaystyle\int (2x+1)^2 \dee{x}$.

Work: We use the substitution $u=2x+1$. Then $\dee{u}=2\dee{x}$, so $\dee{x} = \frac{1}{2}\dee{u}$:

The reasoning is not sound: when we do a substitution, we need to take care of the differential ($\dee{x}$). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here's the way to do it:

Problem: Evaluate $\displaystyle\int (2x+1)^2 \dee{x}$.

Work: We use the substitution $u=2x+1$. Then $\dee{u}=2\dee{x}$, so $\dee{x} = \frac{1}{2}\dee{u}$:

Check the limits.

The problem is with the limits of integration, as in Question 1. Here's how it ought to go:

Problem: Evaluate $\displaystyle\int_{1}^{\pi} \dfrac{\cos(\log t)}{t}\dee{t}$.

Work: We use the substitution $u=\log t$, so $\dee{u}=\frac{1}{t}\dee{t}$. When $t=1$, we have $u=\log 1 =0$ and when $t=\pi$, we have $u=\log(\pi)$. Then:

The problem is with the limits of integration, as in Question 1. Here's how it ought to go:

Problem: Evaluate $\displaystyle\int_{1}^{\pi} \dfrac{\cos(\log t)}{t}\dee{t}$.

Work: We use the substitution $u=\log t$, so $\dee{u}=\frac{1}{t}\dee{t}$. When $t=1$, we have $u=\log 1 =0$ and when $t=\pi$, we have $u=\log(\pi)$. Then:

Check every step. Do they all make sense?

This one is OK.

Perhaps shorter ways exist, but the reasoning here is valid.

Problem: Evaluate $\displaystyle\int_{0}^{\pi/4} x\tan (x^2) \ \dee{x}$.

Work: We begin with the substitution $u=x^2$, $\dee{u} = 2x\dee{x}$:
If $u=x^2$, then $\diff{u}{x} = 2x$, so indeed $\dee{u}=2x\dee{x}$.

Every piece is changed from $x$ to $u$: integrand, differential, limits.

Now we use the substitution $v=\cos u$, $\dee{v}=-\sin u \ \dee{u}$:

Every piece is changed from $u$ to $v$: integrand, differential, limits.

$\displaystyle\int_{0}^{1} \frac{f(u)}{\sqrt{1-u^2}}\,\dee{u}$. Because the denominator $\sqrt{1-u^2}$ vanishes when $u=1$, this is what is known as an improper integral. Improper integrals will be discussed in § 1.12 of the CLP-2 text.

We substitute:

So,

Because the denominator $\sqrt{1-u^2}$ vanishes when $u=1$, this is what is known as an improper integral. Improper integrals will be discussed in § 1.12 of the CLP-2 text.

What is $\diff{}{x}\{f(g(x))\}$?

some constant $C$

Using the chain rule, we see that

So, $\textcolor{red}{f(g(x))}$ is an antiderivative of $\textcolor{red}{f'(g(x))g'(x)}$. All antiderivatives of $f'(g(x))g'(x)$ differ by only a constant, so:

That is, our expression simplifies to some constant $C$.

Remark: since

we conclude

which is precisely how we perform substitution on integrals.

What is the derivative of the argument of the cosine?

$\dfrac{1}{2}\big( \sin(e) - \sin(1) \big)$

We write $\textcolor{red}{u(x) = e^{x^2}}$ and find $\textcolor{blue}{\dee{u} = u'(x)\,\dee{x}=2x e^{x^2}\dee{x}}$. Note that $u(1)=e^{1^2}=e$ when $x=1$, and $u(0)=e^{0^2}=1$ when $x=0$. Therefore:

What is the title of the current section?

$\dfrac{1}{3}$

Substituting $\textcolor{red}{y=x^3}$, $\textcolor{blue}{\dee{y}=3x^2\ \dee{x}}$ :

What is the derivative of $x^3+1$?

$-\dfrac{1}{300{(x^3+1)}^{100}} + C$

Setting $\textcolor{red}{u=x^3+1}$, we have $\textcolor{blue}{\dee{u} = 3x^2\,\dee{x}}$ and so

What is the derivative of $\log x$?

$\log 4$

Setting $\textcolor{red}{u=\log x}$, we have $\textcolor{blue}{\dee{u} = \frac{1}{x}\,\dee{x}}$ and so

since $u=\log(e)=1$ when $x=e$ and $u=\log(e^4)=4$ when $x=e^4$. Then, by the Fundamental Theorem of Calculus Part 2,

What is the derivative of $1+\sin x$?

$\log 2$

Setting $\textcolor{red}{u=1+\sin x}$, we have $\textcolor{blue}{\dee{u} = \cos x~\dee{x}}$ and so

since $u=1+\sin 0=1$ when $x=0$ and $u=1+\sin(\pi/2)=2$ when $x=\pi/2$. Then, by the Fundamental Theorem of Calculus Part 2,

$\cos x$ is the derivative of what?

$\dfrac{4}{3}$

Setting $\textcolor{red}{u=\sin x}$, we have $\textcolor{blue}{\dee{u} = \cos x~\dee{x}}$ and so

since $u=\sin 0=0$ when $x=0$ and $u=\sin(\pi/2)=1$ when $x=\pi/2$. Then, by the Fundamental Theorem of Calculus Part 2,

What is the derivative of the exponent?

$e^6-1$

Substituting $\textcolor{red}{t=x^2-x}$, $\textcolor{blue}{\dee{t} = (2x-1)\,\dee{x}}$ and noting that $t=0$ when $x=1$ and $t=6$ when $x=3$,

What is the derivative of the argument of the square root?

$\dfrac{1}{3}(4-x^2)^{3/2}+C$

We use the substitution $\textcolor{red}{u=4-x^2}$, for which $\textcolor{blue}{\dee{u}=-2x\,\dee{x}}$ :

What is $\diff{}{x}\left\{\sqrt{\log x}\right\}$?

$e^{\sqrt{\log x}}+C$

There is a short, slightly sneaky method — guess an antiderivative — and a really short, still-more-sneaky method.

$0$

Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11 in the

CLP-2 text.

$\dfrac{1}{2}[\cos 1-\cos 2]\approx0.478$

The given sum is of the form

with $\De x=\frac{1}{n}$, $x_j^*=\frac{j}{n}$ and $f(x)=x\sin(1+x^2)$. Since $x_0^*=0$ and $x_n^*=1$, the right hand side is the definition (using the right Riemann sum) of

Using a calculator, we see this is close to $0.478$.

If $w=u^2+1$, then $u^2=w-1$.

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

Often, the denominator of a function is a good guess for the substitution. So, let's try setting $\textcolor{red}{w=u^2+1}$. Then $\textcolor{blue}{\dee{w}=2u\ \dee{u}}$:

The numerator now is $u^2$, and looking at our substitution, we see $\textcolor{red}{u^2=w-1}$:

Using a trigonometric identity, this is similar (though not identical) to $\int \tan \theta \cdot \sec^2 \theta\ \dee{\theta}$.

$\frac{1}{2}\tan^2\theta -\log|\sec \theta|+C$

The only thing we really have to work with is a tangent, so it's worth considering what would happen if we substituted $\textcolor{red}{u=\tan \theta}$. Then $\textcolor{blue}{\dee{u}=\sec^2\theta\ \dee{\theta}}$. This doesn't show up in the integrand as it's written, but we can try and bring it out by using the identity $\tan^2\theta = \sec^2 \theta - 1$:

In Example 1.4.17 of the CLP-2 text, we learned $\int \tan \theta \ \dee{\theta} = \log |\sec \theta|+C$

If you multiply the top and the bottom by $e^x$, what does this look like the antiderivative of?

$\arctan(e^x)+C$

At first glance, it's not clear what substitution to use. If we try the denominator, $u=e^x+e^{-x}$, then $\dee{u}=(e^x-e^{-x})\ \dee{x}$, but it's not clear how to make this work with our integral. So, we can try something else.

If we want to tidy things up, we might think to take $\textcolor{red}{u=e^x}$ as a substitution. Then $\textcolor{blue}{\dee{u}=e^x\ \dee{x}},$ so we need an $e^x$ in the numerator. That can be arranged.

You know methods other than substitution to evaluate definite integrals.

$\dfrac{\pi}{4}-\dfrac{2}{3}$

We often like to take the “inside" function as our substitution, in this case $\textcolor{red}{u=1-x^2}$, so $\textcolor{blue}{\dee{u}=-2x\ \dee{x}}$. This takes care of part of the integral:

The left integral is tough to solve with substitution, but luckily we don't have to–it's the area of a quarter of a circle of radius 1.

$\tan x = \dfrac{\sin x}{\cos x}$

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