## Section2.1Substitution Rule

### Subsection2.1.1Substitution Rule for Indefinite Integrals

Needless to say, most integration problems we will encounter will not be so simple. That is to say we will require more than the basic integration rules we have seen. Here's a slightly more complicated example: Find

\begin{equation*} \int 2x\cos(x^2)\,dx\text{.} \end{equation*}

This is not a “simple” derivative, but a little thought reveals that it must have come from an application of the Chain Rule. Multiplied on the “outside” is $2x\text{,}$ which is the derivative of the “inside” function $\ds x^2\text{.}$ Checking:

\begin{equation*} {d\over dx}\sin(x^2) = \cos(x^2){d\over dx}x^2 = 2x\cos(x^2) \end{equation*}

so

\begin{equation*} \int 2x\cos(x^2)\,dx=\sin(x^2)+C\text{.} \end{equation*}

To summarize: If we suspect that a given function is the derivative of another via the Chain Rule, we let $u$ denote a likely candidate for the inner function, then translate the given function so that it is written entirely in terms of $u\text{,}$ with no $x$ remaining in the expression. If we can integrate this new function of $u\text{,}$ then the antiderivative of the original function is obtained by replacing $u$ by the equivalent expression in $x\text{.}$ This result leads us to the following theorem.

We can describe two methods how the Substitution Rule may unfold in an integration process.

Method 1: Even in simple cases you may prefer to use this mechanical procedure, since it often helps to avoid silly mistakes. For example, consider again this simple problem:

\begin{equation*} \int 2x\cos(x^2)\,dx\text{.} \end{equation*}

Let $\ds u=x^2\text{,}$ then $du/dx = 2x$ or $du = 2x\,dx\text{.}$ Since we have exactly $2x\,dx$ in the original integral, we can replace it by $du\text{:}$

\begin{equation*} \int 2x\cos(x^2)\,dx=\int \cos u\,du=\sin u +C = \sin(x^2)+C\text{.} \end{equation*}

Method 2: This is not the only way to do the algebra, and typically there are many paths to the correct answer. For example, since $du/dx = 2x\text{,}$ we have that $dx=du/2x\text{,}$ and so the integral becomes

\begin{equation*} \int 2x\cos(x^2)\,dx=\int 2x\cos u\,{du\over 2x}=\int \cos u\,du\text{.} \end{equation*}
###### Guideline for Substitution Rule.

Given the integral

\begin{equation*} I=\int f(g(x))g'(x)\,dx \end{equation*}

with $f$ continuous and $g$ differentiable, the following steps outline the Substitution Rule process for integrating $I\text{.}$

1. Let $u=g(x)\text{,}$ which is typically the inside function of the function composition in the integrand.

2. Compute $du=g'(x)dx\text{.}$

3. Write the integral $I$ solely in terms of $u$ using either method:

1. use the substitution $u=g(x)$ and $du=g'(x)dx\text{;}$ or

2. replace $dx$ with $du/g'(x)$ and cancel $g'(x)\text{.}$

4. Integrate with respect to $u\text{.}$

5. Replace $u$ by $g(x)$ to write the result in terms of $x$ only.

Note:

1. However the Substitution Rule unfolds in the solution process, the important part is that you must eliminate all instances of the original variable, often $x\text{,}$ during the process and then write your result back in terms of the original variable $x\text{.}$

2. Sometimes, the integrand has to be rearranged to see whether the Substitution Rule is a possible integration technique.

3. If a first substitution did not work out, then try to simplify or rearrange the integrand to see if a different substitution can be used.

4. As a general guideline for the Substitution Rule, we look for the inside function $u$ to be

• the radicand under a root: e.g.,

\begin{equation*} \text{ when } \int x^3\sqrt{x^2-5}\,dx \text{ we choose } u=x^2-5; \end{equation*}
• the base in a power with a real exponent: e.g.,

\begin{equation*} \text{ when } \int x\left(x^2-5\right)^5\,dx \text{ we choose } u=x^2-5; \end{equation*}
• the exponent in a power with a real base: e.g.,

\begin{equation*} \text{ when } \int x5^{x^2-5}\,dx \text{ we choose } u=x^2-5; \end{equation*}
• the denominator in a fraction: e.g.,

\begin{equation*} \text{ when } \int \frac{x}{x^2-5}\,dx \text{ we choose } u=x^2-5\text{.} \end{equation*}

However, sometimes $u$ can be something different than is suggested above, so be open minded about this process.

###### Example2.2. Substitution Rule.

Evaluate $\ds\int(ax+b)^n\,dx\text{,}$ assuming $a,b$ are constants, $a\not=0\text{,}$ and $n$ is a positive integer.

Solution

We let $u=ax+b$ so $du=a\,dx$ or $dx=du/a\text{.}$ Then

\begin{equation*} \begin{split} \int(ax+b)^n\,dx\amp=\int {1\over a} u^n\,du\\ \amp={1\over a(n+1)}u^{n+1}+C\\ \amp = {1\over a(n+1)}(ax+b)^{n+1}+C. \end{split} \end{equation*}
###### Example2.3. Substitution Rule.

Evaluate $\ds\int \sin(ax+b)\,dx\text{,}$ assuming that $a$ and $b$ are constants and $a\not=0\text{.}$

Solution

Again we let $u=ax+b$ so $du=a\,dx$ or $dx=du/a\text{.}$ Then

\begin{equation*} \begin{split} \int\sin(ax+b)\,dx\amp=\int {1\over a} \sin u\,du\\ \amp={1\over a}(-\cos u)+C\\ \amp= -{1\over a}\cos(ax+b)+C.\end{split} \end{equation*}
###### Example2.4. Substitution in Denominator.

Evaluate the following integral: $\ds\int \frac{2y}{\sqrt{1-4y^2}}\,dy\text{.}$

Solution

We try the substitution:

\begin{equation*} u=1-4y^2\text{.} \end{equation*}

Then,

\begin{equation*} du=-8y~dy \end{equation*}

In the numerator we have $2y~dy\text{,}$ so rewriting the differential gives:

\begin{equation*} -\frac{1}{4}du=2y~dy\text{.} \end{equation*}

Then the integral is:

\begin{equation*} \begin{split} \int \frac{2y}{\sqrt{1-4y^2}}\,dy\amp = \int \left(1-4y^2\right)^{-1/2}(2y~dy)\\ \amp = \int u^{-1/2}\left(-\frac{1}{4}du\right)\\ \amp =\left(\frac{-1}{4}\right)\frac{u^{1/2}}{1/2}+C\\ \amp =-\frac{\sqrt{1-4y^2}}{2}+C\end{split} \end{equation*}
###### Example2.5. Substitution in Base.

Evaluate the following integral: $\ds\int \cos x(\sin x)^5\,dx\text{.}$

Solution

In this question we will let $u=\sin x\text{.}$ Then,

\begin{equation*} du=\cos x~dx\text{.} \end{equation*}

Thus, the integral becomes:

\begin{equation*} \begin{split} \int \cos x(\sin x)^5\,dx\amp =\int u^5\,du\\ =\amp \frac{u^6}{6}+C\\ =\amp \frac{(\sin x)^6}{6}+C\end{split} \end{equation*}
###### Example2.6. Substitution.

Evaluate the following integral: $\ds\int \frac{\cos(\sqrt x)}{\sqrt x}\,dx\text{.}$

Solution

We use the substitution:

\begin{equation*} u=x^{1/2}\text{.} \end{equation*}

Then,

\begin{equation*} du=\frac{1}{2}x^{-1/2}\,dx\text{.} \end{equation*}

Rewriting the differential we get:

\begin{equation*} 2~du=\frac{1}{\sqrt x}~dx\text{.} \end{equation*}

The integral becomes:

\begin{equation*} \begin{split} \int \frac{\cos(\sqrt x)}{\sqrt x}\,dx\amp = 2\int \cos u\,du\\ =\amp 2\sin u+C\\ =\amp 2\sin(\sqrt x)+C\end{split} \end{equation*}
###### Example2.7. Substitution.

Evaluate the following integral: $\ds\int 2x^3\sqrt{x^2+1}\,dx\text{.}$

Solution

This problem is a little bit different than the previous ones. It makes sense to let:

\begin{equation*} u=x^2+1\text{,} \end{equation*}

then

\begin{equation*} du=2x~dx\text{.} \end{equation*}

Making this substitution gives:

\begin{equation*} \begin{split} \int 2x^3\sqrt{x^2+1}\,dx\amp = \int x^2\sqrt{x^2+1}(2x)\,dx\\ =\amp \int x^2u^{1/2}\,du\end{split} \end{equation*}

This is a problem because our integrals can't have a mixture of two variables in them. Usually this means we chose our $u$ incorrectly. However, in this case we can eliminate the remaining $x$'s from our integral by using:

\begin{equation*} u=x^2+1 \implies x^2=u-1\text{.} \end{equation*}

We get:

\begin{equation*} \int x^2u^{1/2}\,du = \int (u-1)u^{1/2}\,du\text{.} \end{equation*}

Now we proceed by simplifying the integrand and noticing we are left with power functions, which are readily integrated.

\begin{equation*} \int (u-1)u^{1/2}\,du=\int u^{3/2}-u^{1/2}\,du=\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}+C\text{.} \end{equation*}

Therefore, our original integral becomes

\begin{equation*} \int 2x^3\sqrt{x^2+1}\,dx=\frac{2}{5}(x^2+1)^{5/2}-\frac{2}{3}(x^2+1)^{3/2}+C \end{equation*}
###### Example2.8. Two Choices for Substitution.

Evaluate $\ds \int \frac{2x}{\sqrt{x^2-5}}\,dx\text{.}$

Solution

Method 1: Let $u=x^2-5\text{,}$ $du=2xdx\text{.}$

\begin{equation*} \begin{split} \int \frac{2x}{\sqrt{x^2-5}}\,dx \amp = \int \frac{du}{u^{1/3}} \\[1ex] \amp = \frac{u^{2/3}}{2/3} + C \\[1ex] \amp = \frac{3}{2} u^{2/3} + C \\[1ex] \amp = \frac{3}{2} \left(x^2-5\right)^{2/3} + C \end{split} \end{equation*}

Method 2: Let $u= \sqrt{x^2-5}\text{,}$ $u^3 = x^2-5\text{,}$ $3u^2du=2xdx\text{.}$

\begin{equation*} \begin{split} \int \frac{2x}{\sqrt{x^2-5}}\,dx \amp = \int \frac{3u^2du}{u} \\[1ex] \amp = 3\int u du\\[1ex] \amp = 3\frac{u^2}{2} + C \\[1ex] \amp = \frac{3}{2} \left(x^2-5\right)^{2/3} + C \end{split} \end{equation*}
###### Example2.9. Application in Sales Projection.

An airline wants to predict the number of tickets on a particular route that will be sold. They find that over the next year, ticket sales growth after $t$ weeks can be modelled by

\begin{equation*} 1000-500e^{-0.025t} \end{equation*}

tickets per week for $t \in [0,52]\text{.}$

1. Determine the predicted total number of tickets sold after the first $t$ weeks.

2. Determine the predicted total number of tickets sold in the next year.

Solution
1. Let $N(t)$ be the total number of tickets predicted to be sold after the first $t$ months. We are given

\begin{equation*} N'(t)= 1000-500e^{-0.025t}\text{.} \end{equation*}

Thus,

\begin{equation*} N(t) = \int \left(1000-500e^{-0.025t}\right)\,dt\text{.} \end{equation*}

We integrate by substitution. Let $u=-0.025t\text{,}$ $du=-0.025 dt\text{:}$

\begin{equation*} \begin{split} N(t) \amp = \int \left(1000-500e^{-0.025t}\right)\,dt \\[1ex] \amp = \int 1000 dt + \frac{500}{0.025} \int e^{u}\,du \\[1ex] \amp = 1000t + 20,000e^{-0.025t} + C \end{split} \end{equation*}

It remains to determine the value of $C\text{.}$ We require

\begin{equation*} N(0) = 0 \implies C=-20,000\text{.} \end{equation*}

Therefore, the total number of tickets sold after the first $t$ months is predicted to be

\begin{equation*} N(t)=1000t+20,000e^{-0.025t}-20,000\text{.} \end{equation*}
2. Since

\begin{equation*} N(52)=1000(52)+20,000e^{-0.025(52)}-20,000 = 37450.64\text{,} \end{equation*}

we have that the model predicts that 37,451 tickets will be sold on this route in the next year.

### Subsection2.1.2Substitution Rule for Definite Integrals

The next example shows how to use the Substitution Rule when dealing with definite integrals.

###### Example2.10. Substitution Rule with Two Methods.

Evaluate $\ds\int_2^4 x\sin(x^2)\,dx\text{.}$

Solution

Method 1: First Compute the Antiderivative, then Evaluate the Definite Integral.

Let $\ds u=x^2\text{,}$ so $du=2x\,dx$ or $x\,dx=du/2\text{.}$ Then

\begin{equation*} \int x\sin(x^2)\,dx=\int {1\over 2} \sin u\,du={1\over 2}(-\cos u)+C= -{1\over 2}\cos(x^2)+C\text{.} \end{equation*}

Now

\begin{equation*} \begin{split} \int_2^4 x\sin(x^2)\,dx\amp=\left.-{1\over 2}\cos(x^2)\right|_2^4\\ \amp =-{1\over 2}\cos(16)+{1\over 2}\cos(4)\end{split}. \end{equation*}

Method 2: Changing the Limits of Integration with the Substitution.

A somewhat neater alternative to this method is to change the original limits to match the variable $u\text{.}$ Since $\ds u=x^2\text{,}$ when $x=2\text{,}$ $u=4\text{,}$ and when $x=4\text{,}$ $u=16\text{.}$ So we can do this:

\begin{equation*} \begin{split} \int_2^4 x\sin(x^2)\,dx \amp= \int_4^{16} {1\over 2} \sin u\,du\\ \amp =\left.-{1\over 2}(\cos u)\right|_4^{16} \\ \amp =-{1\over 2}\cos(16)+{1\over 2}\cos(4).\end{split} \end{equation*}

An incorrect, and dangerous, alternative is something like this:

\begin{equation*} \begin{split} \int_2^4 x\sin(x^2)\,dx\amp =\int_2^4 {1\over 2} \sin u\,du\\ \amp= \left.-{1\over 2}\cos (u)\right|_2^4\\ \amp= \left.-{1\over 2}\cos(x^2)\right|_2^4\\ \amp =-{1\over 2}\cos(16)+{1\over 2}\cos(4).\end{split} \end{equation*}

This is incorrect because $\ds\int_2^4 {1\over 2} \sin u\,du$ means that $u$ takes on values between 2 and 4, which is wrong. It is dangerous, because it is very easy to get to the point $\ds\left.-{1\over 2}\cos (u)\right|_2^4$ and forget to substitute $\ds x^2$ back in for $u\text{,}$ thus getting the incorrect answer $\ds -{1\over 2}\cos(4)+{1\over 2}\cos(2)\text{.}$ An acceptable alternative is to clearly indicate that the limit substitution is to be done with respect to the $x$ variable, using something like:

\begin{equation*} \begin{split} \int_2^4 x\sin(x^2)\,dx\amp=\int_{x=2}^{x=4} {1\over 2} \sin u\,du\\ \amp = \left.-{1\over 2}\cos (u)\right|_{x=2}^{x=4}\\ \amp= \left.-{1\over 2}\cos(x^2)\right|_2^4\\ \amp=-{\cos(16)\over 2}+{\cos(4)\over2}.\end{split} \end{equation*}

To summarize, we have the following.

Note: When using the Substitution Rule for integrating definite integrals, it is important to change the limits of integration from those of the original function to those of the substituted function. Otherwise, the definite integral will evaluate to an incorrect result. ###### Example2.12. Substitution Rule for Definite Integrals.

Evaluate $\ds\int_{1/4}^{1/2}{\cos(\pi t)\over\sin^2(\pi t)}\,dt\text{.}$

Solution

Let $u=\sin(\pi t)$ so $du=\pi\cos(\pi t)\,dt$ or $du/\pi=\cos(\pi t)\,dt\text{.}$ We change the limits to $\ds u(1/4)=\sin(\pi/4)=\sqrt2/2$ and $u(1/2)=\sin(\pi/2)=1\text{.}$ Then

\begin{equation*} \begin{split} \int_{1/4}^{1/2}{\cos(\pi t)\over\sin^2(\pi t)}\,dt \amp= \int_{\sqrt2/2}^{1}{1\over \pi}{1\over u^2}\,du\\ \amp= \int_{\sqrt2/2}^{1} {1\over \pi}u^{-2}\,du\\ \amp = \left.{1\over \pi}{u^{-1}\over -1}\right|_{\sqrt2/2}^{1}\\ \amp= -{1\over\pi}+{\sqrt2\over\pi}.\end{split} \end{equation*}
###### Example2.13. Substitution Rule for Definite Integrals.

Evaluate $\ds\int_{-1}^{1} \left(y+1\right)\left(y^2+2y\right)^8dy\text{.}$

Solution

Although we could expand the integrand, since this would yield powers of $y$ which we can certainly integrate without using the Substitution Rule at all, the exponent 8 would make this a rather messy process that is surely prone to errors. Instead we proceed with the obvious choice of substitution and let $u=y^2+2y\text{,}$ then

\begin{equation*} dy=(2y+2)dy = 2(y+1)dy \implies \frac{du}{2}=(y+1)dy \end{equation*}

with limits of integration

\begin{equation*} u(-1)=(-1)^2+2(-1)=-1 \text{ and } u(1)=(1)^2+2(1)=3\text{.} \end{equation*}

Take note that sometimes the value of a limit of integration does not change. The point is that one still needs to substitute the values of the original variable of integration, in this case $y\text{,}$ to work with the limits of integration for the substituted variable, in this case $u\text{.}$

Then the integral evaluates as follows

\begin{equation*} \begin{split} \int_{-1}^{1} \left(y+1\right)\left(y^2+2y\right)^8dy \amp= \int_{-1}^{3}\frac{u^8}{2}du\\ \amp = \frac{u^9}{18}\bigg\vert_{-1}^3\\ \amp= \frac{3^9}{18}-\frac{(-1)^9}{18}\\ \amp = \frac{19,684}{18}.\end{split} \end{equation*}
##### Exercises for Section 2.1.

Evaluate the following indefinite integrals.

1. $\ds\int (1-t)^9\,dt$

$\ds -(1-t)^{10}/10+C$
Solution

Let $u=1-t\text{.}$ Then $du = -dt$ and

\begin{equation*} \int (1-t)^9\,dt = -\int u^9 \,du = -\frac{1}{10} u^{10} + C = -\frac{(1-t)^{10}}{10} + C\text{.} \end{equation*}
2. $\ds\int (x^2+1)^2\,dx$

$\ds x^5/5+2x^3/3+x+C$
Solution

First, try $u=x^2+1\text{.}$ Then $du = 2x\,dx$ and

\begin{equation*} \int (x^2+1)^2\,dx = \int u^2 \frac{du}{2x}\text{,} \end{equation*}

and so this does choice of substitution will not work. Instead, we solve directly:

\begin{equation*} \int (x^2+1)^2\,dx = \int (x^4+2x^2+1)\,dx = \frac{x^5}{5}+\frac{2x^3}{3}+x+C\text{.} \end{equation*}
3. $\ds\int x(x^2+1)^{100}\,dx$

$\ds (x^2+1)^{101}/202+C$
Solution

Let $u=x^2+1\text{.}$ Then $du = 2x\,dx$ and

\begin{equation*} \int x(x^2+1)^{100}\,dx = \int xu^{100} \, \frac{du}{2x} = \frac{1}{2} \int u^{100}\,du\text{.} \end{equation*}

This is now an integral we can solve:

\begin{equation*} \int u^{100}\,du = \frac{1}{101} u^{101} + C\text{,} \end{equation*}

and so we have

\begin{equation*} \int x(x^2+1)^{100}\,dx = \frac{1}{202} (x^2+1)^{101} + C\text{.} \end{equation*}
4. $\ds\int {1\over\root 3 \of {1-5t}}\,dt$

$\ds -3(1-5t)^{2/3}/10+C$
Solution

Let $u=1-5t\text{.}$ Then $du= -5\,dt$ and

\begin{equation*} \int \frac{1}{\sqrt{1-5t}}\,dt = \int \frac{1}{-5\sqrt{u}}\,du\text{.} \end{equation*}

We now carry out the integration with respect to $u\text{,}$ and then rewrite our answer in terms of $t\text{:}$

\begin{equation*} \int \frac{1}{-5\sqrt{u}}\,du = \frac{-3}{10}u^{2/3}+C = \frac{-3}{10}(1-5t)^{2/3}+C\text{.} \end{equation*}
5. $\ds\int \sin^3x\cos x\,dx$

$\ds (\sin^4x)/4+C$
Solution

Let $u=\sin x\text{.}$ Then $du = \cos x\,dx$ and

\begin{equation*} \begin{split} \int \sin^3 x\cos x\,dx \amp = \int u^3 \,du\\ \amp = \frac{1}{4} u^4 + C\\ \amp = \frac{1}{4} \sin^4x + C. \end{split} \end{equation*}
6. $\ds\int x\sqrt{100-x^2}\,dx$

$\ds -(100-x^2)^{3/2}/3+C$
Solution

Let $u=100-x^2\text{.}$ Then $du = -2x\,dx$ and

\begin{equation*} \begin{split} \int x \sqrt{100-x^2}\,dx \amp = -\frac{1}{2}\int \sqrt{u}\,du \\ \amp = -\frac{1}{3} u^{3/2} + C\\ \amp = -\frac{1}{3} \left(100-x^2\right)^{3/2} + C. \end{split} \end{equation*}
7. $\ds\int {x^2\over\sqrt{1-x^3}}\,dx$

$\ds \ds -2\sqrt{1-x^3}/3+C$
Solution
Let $u=1-x^3\text{,}$ with $du=-3x^2\text{:}$
\begin{equation*} \begin{split} \int \frac{x^2}{\sqrt{1-x^3}}\,dx \amp = -\frac{1}{3}\int \frac{1}{\sqrt{u}}\,du\\ \amp = -\frac{2}{3} \sqrt{u} + C\\ \amp= -\frac{2}{3}\sqrt{1-x^3} + C.\end{split} \end{equation*}
8. $\ds\int \cos(\pi t)\cos\bigl(\sin(\pi t)\bigr)\,dt$

$\ds \sin(\sin\pi t)/\pi+C$
Solution

Let $u=\sin(\pi t)\text{.}$ Then $du = \pi\cos(\pi t)$ and

\begin{equation*} \begin{split} \int \cos(\pi t)\cos(\sin(\pi t))\,dt \amp = \frac{1}{\pi} \int \cos u \,du \\ \amp = \frac{1}{\pi} \sin u + C\\ \amp = \frac{1}{\pi} \sin(\sin(\pi t)) + C. \end{split} \end{equation*}
9. $\ds\int {\sin x\over\cos^3 x}\,dx$

$\ds \ds 1/(2\cos^2 x)=(1/2)\sec^2x+C$
Solution

Let $u=\cos x\text{.}$ Then $du = -\sin x\,dx$ and

\begin{equation*} \begin{split} \int \frac{\sin x}{\cos^3 x}\,dx \amp = -\int \frac{1}{u^3}\,du \\ \amp = \frac{1}{2 u^2} + C\\ \amp = \frac{1}{2 \cos^2 x} + C. \end{split} \end{equation*}
10. $\ds\int\tan x\,dx$

$-\ln|\cos x|+C$
Solution

Let $u=\cos x\text{.}$ Then $du=-\sin x\,dx$ and

\begin{equation*} \begin{split} \int \frac{\sin x}{\cos x}\,dx \amp = -\int \frac{1}{u}\,du \\ \amp = -\ln|u|+C\\ \amp = -\ln|\cos x| + C. \end{split} \end{equation*}
11. $\ds\int\sec^2x\tan x\,dx$

$\ds \sec^2(x)/2+C$
Solution

Let $u=\sec x\,dx\text{.}$ Then $du = \tan x \sec x\,dx$ and

\begin{equation*} \begin{split} \int \sec^2 x\tan x\,dx \amp = \int u\,du\\ \amp =\frac{1}{2} u^2 + C\\ \amp =\frac{1}{2} \sec^2x + C. \end{split} \end{equation*}
12. $\ds\int {\sin(\tan x)\over\cos^2x}\,dx$

$-\cos(\tan x)+C$
Solution

Let $u=\tan x\text{.}$ Then $du = \sec^2 x\,dx$ and

\begin{equation*} \begin{split} \int \frac{\sin(\tan x)}{\cos^2 x}\,dx \amp = \int \sin u\,du\\ \amp = -\cos u + C\\ \amp = -\cos(\tan x) + C. \end{split} \end{equation*}
13. $\ds\int {6x\over(x^2 - 7)^{1/9}}\,dx$

$\ds (27/8)(x^2-7)^{8/9}$
Solution

Let $u = x^2-7\text{.}$ Then $du = 2x\,dx$ and

\begin{equation*} \begin{split} \int \frac{6x}{(x^2-7)^{1/9}} \,dx \amp = \int \frac{3}{u^{1/9}}\,du\\ \amp = \frac{27}{8} u^{8/9} + C\\ \amp = \frac{27}{8} (x^2-7)^{8/9}. \end{split} \end{equation*}
14. $\ds\int f(x) f'(x)\,dx$

$\ds f(x)^2/2$
Solution

Let $u = f(x)\text{.}$ Then $du = f'(x)\,dx$ and

\begin{equation*} \begin{split} \int f(x) f'(x)\,dx \amp = \int u\,du\\ \amp = \frac{1}{2} u^2 + C\\ \amp = \frac{1}{2} \left[f(x)\right]^2 + C. \end{split} \end{equation*}

Evaluate the following definite integrals.

1. $\ds\int_0^\pi\sin^5(3x)\cos(3x)\,dx$

$0$
Solution

Let $u = \sin(3x)\text{.}$ Then $du = 3\cos(3x)\,dx\text{,}$ and the integration bounds are from

\begin{equation*} \sin(0) = 0 \text{ to } \sin(3\pi) = 0\text{.} \end{equation*}

Therefore, we have

\begin{equation*} \int_0^\pi \sin^5(3x)\cos(3x)\,dx = \int_0^0 u^5 \left(\frac{du}{3}\right) = 0\text{.} \end{equation*}
2. $\ds\int_0^{\sqrt{\pi}/2} x\sec^2(x^2)\tan(x^2)\,dx$

$1/4$
Solution

We know that

\begin{equation*} \diff{}{x} \sec(x) = \tan(x)\sec(x)\text{,} \end{equation*}

so take the substitution

\begin{equation*} u=\sec(x^2) \implies du = 2x\sec(x^2)\tan(x^2)\,dx\text{.} \end{equation*}

When $x=0\text{,}$ we have that $u=\sec(0) = 1\text{,}$ and when $x=\sqrt{\pi}/2\text{,}$ $u=\sec(\pi/4)= \sqrt{2}\text{.}$ And so in terms of $u\text{,}$ the integral becomes

\begin{equation*} \int_1^{\sqrt{2}} u \left(\frac{du}{2}\right) = \frac{1}{2} \left[\frac{1}{2}u^2\right]_1^{\sqrt{2}} = \frac{1}{2}-\frac{1}{4} = \frac{1}{4}\text{.} \end{equation*}
3. $\ds\int_3^4 {1\over(3x-7)^2}\,dx$

$1/10$
Solution

Take $u=3x-7$ with $du=3\,dx\text{.}$ The integration bounds are thus

\begin{equation*} x=3 \implies u = 9-7 = 2, \text{ and } x=4 \implies u = 12-7 = 5\text{.} \end{equation*}

We now compute

\begin{equation*} \displaystyle \int_3^4 \frac{1}{(3x-7)^2}\,dx = \int_2^5 \frac{1}{u^2} \left(\frac{du}{3}\right) = \frac{1}{3}\left[-\frac{1}{u}\right]_2^5 = \frac{1}{10}\text{.} \end{equation*}
4. $\ds\int_0^{\pi/6}(\cos^2x - \sin^2x)\,dx$

$\ds \sqrt3/4$
Solution

We first use a trigonometric identity to simplify the integrand:

\begin{equation*} \int_0^{\pi/6} \left(\cos^2 x- \sin^2x\right)\,dx = \int_0^{\pi/6} \cos (2x)\,dx\text{.} \end{equation*}

Now let $u=2x$ and $du = 2\,dx\text{.}$ Then the integral becomes

\begin{equation*} \int_0^{\pi/3} \frac{\cos u}{2}\,du = \frac{\sin u}{2} \bigg\vert_0^{\pi/3} = \frac{\sqrt{3}}{4}\text{.} \end{equation*}
5. $\ds\int_{-1}^1 (2x^3-1)(x^4-2x)^6\,dx$

$\ds -(3^7+1)/14$
Solution

Let $u=x^4-2x\text{.}$ Then $du = 4x^3-2 = 2(2x^3-1)\text{.}$ Therefore, when $x=-1\text{,}$ $u=1+2=3\text{,}$ and when $x=1\text{,}$ $u=1-2=-1\text{.}$ The integral then becomes

\begin{equation*} \begin{split} \int_{-1}^1 (2x^3-1)(x^4-2x)^6\,dx \amp = -\frac{1}{2}\int_{-1}^{3} u^6\,du \\ \amp =\frac{-1}{14} u^7 \bigg\vert_{-1}^3\\ \amp = \frac{-1}{14} \left(3^7 +1\right)\\ \amp = -\frac{1094}{7}. \end{split} \end{equation*}
6. $\ds\int_{-1}^1 \sin^7 x\,dx$

$0$
Solution

Let $u=\cos x$ and $du = -\sin x\,dx\text{.}$ Then:

\begin{equation*} \int_{-1}^1 \sin^7 x\,dx = -\int_{\cos(-1)}^{\cos(1)} \left(1-u^2\right)^3\,du = 0\text{.} \end{equation*}

A toy truck manufacturer estimates that the number of sales after Christmas declines at a rate of $-3e^{-0.2t}$ toys per day ($t \in [0,60]$). If the manufacturer sells $10,000$ units on Christmas day ($t=0$), determine the number of expected sales after $t$ days.

$15e^{-0.2t}+9985$ trucks.
Solution

Let $S(t)$ denote the number of sales after $t$ days after Christmas. Then we know that $S'(t) = -3 e^{-0.2t}$ and $S(0) = 10,000\text{.}$ Therefore,

\begin{equation*} S(t) = -3\int e^{-0.2t}, dt\text{.} \end{equation*}

Let $u=-0.2t\text{,}$ and $du=-0.2\,dt\text{.}$ Then,

\begin{equation*} \int e^{-0.2t}\,dt = -\frac{10}{2}\int e^{u}\,du = -\frac{10}{2} e^{u}+C\text{.} \end{equation*}

So we must have

\begin{equation*} S(t) = 15 e^{-0.2t} +C\text{.} \end{equation*}

Now using the initial condition, we find

\begin{equation*} S(0) = 15+C = 10,000 \implies C = 9985\text{.} \end{equation*}

All together, we find that

\begin{equation*} S(t) = 15e^{-0.2t}+9985\text{.} \end{equation*}

Let $q$ be the quantity (in thousands) of a product in the market when the unit price is set at $p$ dollars per unit. There are currently $2000$ units available at a price of $2/unit. Determine the corresponding supply equation if the price increases at a rate of \begin{equation*} p'(q)=\frac{100q}{(1-q)^2}\text{.} \end{equation*} Answer $p(q)= 100(1/(1-q)+\log(q-1))+102$ Solution We first solve the indefinite integral: \begin{equation*} \int p'(q)\,dq = \int \frac{100q}{(1-q)^2}\,dq\text{.} \end{equation*} Let $u=1-q$ and $du = -dq\text{.}$ Then $q=1-u$ and \begin{equation*} \begin{split} 100 \int \frac{q}{(1-q)^2}\,dq \amp = -100 \int \frac{1-u}{u^2}\,du\\ \amp = -100\left(\int \frac{1}{u^2}\,du - \int \frac{1}{u}\,du\right)\\ \amp = -100\left(\frac{-1}{u} -\ln |u| \right) + C\\ \amp = \frac{100}{1-q} +100\ln|1-q|+C \end{split} \end{equation*} Therefore, for some constant $C\text{,}$ we have \begin{equation*} p(q) = \frac{100}{1-q} +100\ln|1-q|+C\text{.} \end{equation*} Since 2000 units ($q=2$) are available at$2 per unit, we know that

\begin{equation*} 2 = p(2) = \frac{100}{-1} + 100 \ln (1)+ C = C-100 \implies C = 102\text{.} \end{equation*}

All together, we find that the supply equation is

\begin{equation*} p(q) = \frac{100}{1-q} +100\ln|1-q|+102\text{.} \end{equation*}

Let $q$ represent the daily quantity demanded (in thousands) of a certain product. When the unit price $p$ is set at \$$3\text{,}$ it is observed that $1000$ units are demanded per day. Determine the corresponding demand equation if the price decreases at a rate of

\begin{equation*} p'(q)=\frac{-200q}{(3+q^2)^{3/2}}\text{.} \end{equation*}
$p(q)=\frac{200}{\sqrt{q^2+3}}-97$
Solution

We are given that a certain demand equation decreases at a rate of

\begin{equation*} p'(q) = \frac{-200q}{(3+q^2)^{3/2}} \end{equation*}

dollars per day, where $q$ is the number of units demanded per day, measured in thousands. To find the corresponding demand function, we first set up the indefinite integral,

\begin{equation*} p(q) = \int p'(q)\,dq = \int \frac{-200q}{(3+q^2)^{3/2}} \,dq\text{.} \end{equation*}

We solve by substitution: Let

\begin{equation*} u=3+q^2, \ du = 2q\,dq\text{.} \end{equation*}

Thus,

\begin{equation*} p(u) = \int \frac{-200}{u^{3/2}} \frac{du}{2} = \frac{200}{\sqrt{u}} + C\text{.} \end{equation*}

So in terms of $q\text{,}$

\begin{equation*} p(q) = \frac{200}{\sqrt{3+q^2}} + \hat{C}\text{.} \end{equation*}

We are further given that $p(1) = 3\text{,}$ so we must have that

\begin{equation*} \frac{200}{\sqrt{4}} + \hat{C} = 3 \implies \hat{C}=-97\text{.} \end{equation*}

The demand function of the product is thus

\begin{equation*} p(q) = \frac{200}{\sqrt{3+q^2}} - 97\text{.} \end{equation*}

In the first month ($t=0$) following a successful marketing campaign, a company sells 2,000 units of their product. Management predicts that the sales will decline at a rate of

\begin{equation*} N'(t)=-200(e^{-2t}+1) \end{equation*}

units per month for the next 3 months. Determine the total number of expected sales after 3 months.

Approximately 1300 units.
Solution

We first integrate

\begin{equation*} \int -200 \left(e^{-2t}+1\right)\,dt \end{equation*}

Let $u=-2t$ and $du = -2\,dt\text{.}$ Then:

\begin{equation*} \begin{split} \int -200 \left(e^{-2t}+1\right)\,dt \amp = 100\int e^u\,du + 100\int\,du \\ \amp = 100 e^u + 100 u + C\\ \amp = 100 e^{-2t}-200 t + C \end{split} \end{equation*}

Therefore, if 2,000 units were sold in the first month, we must have

\begin{equation*} 2000 = N(0) = 100+C \implies C =1900\text{.} \end{equation*}

Hence, the number of units sold can be described by the equation

\begin{equation*} N(t) = -200 e^{-2t} - 200t + 1900\text{.} \end{equation*}

After three months, the number of units sold will then be

\begin{equation*} N(3) = -200 e^{-6} - 600 + 1900 = 1299.50\text{,} \end{equation*}

or about 1300 units.