## Section2.2Powers of Trigonometric Functions

The trigonometric substitutions we will focus on in this section are summarized in the table below:

\begin{equation*} \begin{array}{lllll} \textbf{ Substitution} \amp u=\sin x \amp u=\cos x \amp u=\tan x \amp u=\sec x\\[0.5em] \textbf{ Derivative} \amp du=\cos x\, dx \amp du=-\sin x\,dx \amp du=\sec^2x\,dx \amp du=\sec x\tan x\,dx \end{array} \end{equation*}

### Subsection2.2.1Products of Powers of Sine and Cosine

Functions consisting of powers of the sine and cosine can be integrated by using substitution and trigonometric identities. These can sometimes be tedious, but the technique is straightforward. A similar technique is applicable to powers of secant and tangent as shown in Section 2.2.2 (and also cosecant and cotangent, not discussed here). An example will suffice to explain the approach.

###### Example2.14. Odd Power of Sine.

Evaluate $\ds\int \sin^5 x\,dx\text{.}$

Solution

Rewrite the function:

\begin{equation*} \begin{split} \int \sin^5 x\,dx\amp = \int \sin x \sin^4 x\,dx\\ =\amp \int \sin x (\sin^2 x)^2\,dx\\ =\amp \int \sin x (1-\cos^2 x)^2\,dx\end{split}\text{.} \end{equation*}

Now use $u=\cos x\text{,}$ $du=-\sin x\,dx\text{:}$

\begin{equation*} \begin{split} \int \sin x (1-\cos^2 x)^2\,dx=\amp \int -(1-u^2)^2\,du\\ =\amp \int -(1-2u^2+u^4)\,du\\ =\amp \int 1+2u^2-u^4\,du\\ =\amp -u+{\frac{2}{3}}u^3-{\frac{1}{5}}u^5+C\\ =\amp -\cos x+\frac{2}{3}\cos^3 x-\frac{1}{5}\cos^5x+C\end{split}\text{.} \end{equation*}

Observe that by taking the substitution $u=\cos x$ in the last example, we ended up with an even power of sine from which we can use the formula $\sin^2x+\cos^2x=1$ to replace any remaining sines. We then ended up with a polynomial in $u$ in which we could expand and integrate quite easily. Notice here that the “obvious” substitution $u=\sin x$ in the original integral does not lead to any useful simplification.

This technique works for products of powers of sine and cosine. We summarize it below.

###### Guideline for Integrating Products of Sine and Cosine.

When evaluating $\ds\int\sin^mx \cos^nx\,dx\text{:}$

1. The power of sine is odd ($m$ odd):

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

2. Replace $dx$ using (a), thus cancelling one power of $\sin x$ by the substitution of $du\text{,}$ and be left with an even number of sine powers.

3. Use $\sin^2x=1-\cos^2x~(=1-u^2)$ to replace the leftover sines.

2. The power of cosine is odd ($n$ odd):

1. Use $u=\sin x$ and $du=\cos x\,dx\text{.}$

2. Replace $dx$ using (a), thus cancelling one power of $\cos x$ by the substitution of $du\text{,}$ and be left with an even number of cosine powers.

3. Use $\cos^2x=1-\sin^2x~(=1-u^2)$ to replace the leftover cosines.

3. Both $m$ and $n$ are odd: Use either $1$ or $2$ (both will work).

4. Both $m$ and $n$ are even: Use $\cos^2x=\frac{1}{2}\left(1+\cos(2x)\right)$ and/or $\sin^2x=\frac{1}{2}\left(1-\cos(2x)\right)$ to reduce to a form that can be integrated.

Note: As $m$ and $n$ get large, multiple steps will be needed.
###### Example2.15. Odd Power of Cosine and Even Power of Sine.

Evaluate $\ds\int \sin^6 x\cos^5 x\,dx\text{.}$

Solution

We will show this solution in two ways.

Method 1: We rewrite the integrand to use the Substitution Rule.

\begin{equation*} \begin{split} \int \sin^6 x\cos^5 x\,dx \amp = \int \sin^6 x \cos^4 x \cos x\,dx\\ =\amp \int \sin^6 x (\cos^2 x)^2 \cos x\,dx\\ =\amp \int \sin^6 x (1-\sin^2 x)^2 \cos x\,dx\end{split} \end{equation*}

Then use the substitution $u=\sin x$ since the derivative of $\sin x$ is $\cos x\text{,}$ and also because there is only one occurrence of cosine in the integrand. Hence, $du=\cos x\,dx$ is the perfect substitution.

\begin{equation*} \begin{split} \int \sin^6 x (1-\sin^2 x)^2 \cos x\,dx \amp = \int u^6 (1-u^2)^2 du\\ =\amp \int u^6-2u^8+u^{10}\,du\\ =\amp \frac{u^7}{7}-\frac{2u^9}{9}+\frac{u^{11}}{11}+C\\ =\amp \frac{\sin^7x}{7}-\frac{2\sin^9x}{9}+\frac{\sin^{11}x}{11}+C\end{split} \end{equation*}

Method 2: We apply the Guideline for Integrating Products of Sine and Cosine. Since the power of cosine is odd, we use the substitution $u=\sin x$ and $du=\cos x\text{.}$

Then

\begin{align*} \int \sin^6 x\cos^5 x\,dx \amp = \int u^6\cos^5x \,\ds\frac{du}{\cos x} \amp \amp \mbox{Using the substitution}\\ \amp = \int u^6\left(\cos^2x\right)^2 \,du \amp \amp \mbox{Canceling a $\cos x$ and rewriting $\cos^4x$}\\ \amp = \int u^6(1-\sin^2x)^2\,du \amp \amp \mbox{Using trig identity $\cos^2x=1-\sin^2x$}\\ \amp = \int u^6(1-u^2)^2\,du \amp \amp \mbox{Writing integral in terms of $u$'s}\\ \amp = \int u^6-2u^8+u^{10}\,du \amp \amp \mbox{Expand and collect like terms}\\ \amp = \frac{u^7}{7}-\frac{2u^9}{9}+\frac{u^{11}}{11}+C \amp \amp \mbox{Integrating}\\ \amp = \frac{\sin^7x}{7}-\frac{2\sin^9x}{9}+\frac{\sin^{11}x}{11}+C \hspace{3ex} \amp \amp \mbox{Replacing $u$ back in terms of $x$} \end{align*}
###### Example2.16. Odd Power of Cosine.

Evaluate $\ds\int \cos^3 x\,dx\text{.}$

Solution

Since the power of cosine is odd, we use the substitution $u=\sin x$ and $du=\cos x\,dx\text{.}$ This may seem strange at first since we don't have $\sin x$ in the question, but it does work!

\begin{align*} \int \cos^3 x\,dx \amp = \int \cos^3x \,\ds\frac{du}{\cos x} \amp\amp \mbox{Using the substitution} \\ \amp = \int \cos^2x \,du \amp \amp \mbox{Canceling a $\cos x$} \\ \amp = \int (1-\sin^2x)\,du \amp\amp \mbox{Using trig identity $\cos^2x=1-\sin^2x$} \\ \amp = \int (1-u^2)\,du \amp\amp \mbox{Writing integral in terms of $u$'s} \\ \amp = u-\frac{u^3}{3}+C \amp \amp \mbox{Integrating} \\ \amp = \sin x-\frac{\sin^3x}{3}+C \amp \amp \mbox{Replacing $u$ back in terms of $x$} \end{align*}
###### Example2.17. Product of Even Powers of Sine and Cosine.

Evaluate $\ds\int\! \sin^2x\cos^2x\,dx\text{.}$

Solution

Use the formulas

\begin{equation*} \ds \sin^2x = \frac{1-\cos(2x)}{2} \text{ and } \cos^2x =\frac{1+\cos(2x)}{2} \end{equation*}

to get

\begin{equation*} \int \sin^2x\cos^2x\,dx=\int {1-\cos(2x)\over2}\cdot {1+\cos(2x)\over2}\,dx\text{.} \end{equation*}

We then have

\begin{align*} \int \sin^2x\cos^2x\,dx\amp = \int {1-\cos(2x)\over2}\cdot{1+\cos(2x)\over2}\,dx\\ \amp ={1\over4}\int 1-\cos^2 2x\,dx\\ \amp = {1\over4}\left(x-\int\cos^2 2x\,dx\right)\text{.} \end{align*}

To continue the integration, we use the cosine double angle identity again with

\begin{equation*} \cos^2(2x) = \frac{1+\cos(4x)}{2}\text{.} \end{equation*}

Then

\begin{align*} \int \sin^2x\cos^2x\,dx\amp ={1\over4}\left(x-{1\over2}\int 1+\cos4x \,dx\right)\\ \amp = {1\over4}\left(x-{1\over2}\left(x+{\sin 4x\over 4}\right)\right)\\ \amp = {1\over4}\left(x-{x\over2}-{\sin 4x\over 8}\right)+C \end{align*}
###### Example2.18. Even Power of Sine.

Evaluate $\ds\int \sin^6 x\,dx\text{.}$

Solution

Use $\ds \sin^2x =(1-\cos(2x))/2$ to rewrite the function:

\begin{align*} \int \sin^6 x\,dx\amp = \int (\sin^2 x)^3\,dx\\ \amp = \int {(1-\cos 2x)^3\over 8}\,dx\\ \amp ={1\over 8}\int 1-3\cos 2x+3\cos^2 2x-\cos^3 2x\,dx\text{.} \end{align*}

Now we have four integrals to evaluate. Ignoring the constant for now:

\begin{equation*} \int 1\,dx=x \end{equation*}

and

\begin{equation*} \int -3\cos 2x\,dx = -{3\over 2}\sin 2x \end{equation*}

are easy. The $\ds \cos^3 2x$ integral is like the previous example:

\begin{align*} \int -\cos^3 2x\,dx\amp = \int -\cos 2x\cos^2 2x\,dx\\ \amp = \int -\cos 2x(1-\sin^2 2x)\,dx\\ \amp = \int -{1\over 2}(1-u^2)\,du\\ \amp = -{1\over 2}\left(u-{u^3\over 3}\right)\\ \amp = -{1\over 2}\left(\sin 2x-{\sin^3 2x\over 3}\right)\text{.} \end{align*}

And finally we use another trigonometric identity, $\ds \cos^2x=(1+\cos(2x))/2\text{:}$

\begin{equation*} \int 3\cos^2 2x\,dx=3\int {1+\cos 4x\over 2}\,dx= {3\over 2}\left(x+{\sin 4x\over 4}\right)\text{.} \end{equation*}

So at long last we get

\begin{equation*} \int \sin^6 x\,dx = {x\over8} -{3\over 16}\sin 2x -{1\over 16}\left(\sin 2x-{\sin^3 2x\over 3}\right) +{3\over 16}\left(x+{\sin 4x\over 4}\right)+C\text{.} \end{equation*}

### Subsection2.2.2Exploring Powers of Secant and Tangent

Now, we turn our attention to products of secant and tangent. Some we already know how to do from the table of Integral Rules in Section 1.5.3.

Note: A common mistake is to believe that $\int\tan x\,dx$ is $\sec^2(x)+C$ — this is not true. We can readily integrate $\tan x$ to demonstrate why this is not true.

###### Example2.19. Integrating Tangent.

Evaluate $\ds\int\tan x\,dx\text{.}$

Solution

Note that $\ds\tan x = \frac{\sin x}{\cos x}$ and let $u=\cos x\text{,}$ so that $du=-\sin x\,dx\text{.}$

\begin{align*} \int \tan x\,dx \amp = \int \frac{\sin x}{\cos x}\,dx \amp\amp \mbox{Rewriting $\tan x$} \\ \amp = \int \frac{\sin x}{u}\,\ds\frac{du}{-\sin x} \amp\amp \mbox{Using the substitution}\\ \amp = -\int \frac{1}{u}\,du \amp \amp \mbox{Cancelling and pulling the $-1$ out} \\ \amp = -\ln|u|+C \amp \amp \mbox{Using formula $\ds\int\frac{1}{u}\,dx=\ln|u|+C$} \\ \amp = -\ln|\cos x|+C \amp\amp \mbox{Replacing $u$ back in terms of $x$} \\ \amp = \ln|\sec x|+C \amp \amp \mbox{Using log properties and $\sec x=1/\cos x$} \end{align*}
###### Example2.20. Integrating Tangent Squared.

Evaluate $\ds\int\tan^2 x\,dx\text{.}$

Solution

Note that $\ds\tan^2 x = \sec^2x-1\text{.}$

\begin{align*} \ds\int \tan^2 x\,dx \amp = \ds\int \left(\sec^2x-1\right)\,dx \amp\amp \mbox{Rewriting $\tan x$} \\ \amp = \tan x-x+C \amp \amp \mbox{Since $\ds\int\sec^2x\,dx=\tan x+C$} \end{align*}

Note: In problems with tangent and secant, two integrals come up frequently:

Both have relatively nice expressions but they are a bit tricky to discover.

First we evaluate $\ds\int\sec x\,dx\text{,}$ which we will need to compute $\ds \int\sec^3x\,dx\text{.}$

###### Example2.21. Integral of Secant.

Evaluate $\ds\int\sec x\,dx\text{.}$

Solution

We start with a trick, namely we multiply the integrand by 1, but we express 1 as the ratio $(\sec x + \tan x)/(\sec x + \tan x)\text{.}$ This sounds like a crazy idea, but it works!

\begin{align*} \int\sec x\,dx\amp = \int\sec x\,{\sec x +\tan x\over \sec x +\tan x}\,dx\\ \amp = \int{\sec^2 x +\sec x\tan x\over \sec x +\tan x}\,dx\text{.} \end{align*}

Now let $u=\sec x +\tan x\text{,}$ $\ds du=\sec x \tan x + \sec^2x\,dx\text{,}$ exactly the numerator of the function we are integrating. Thus

\begin{align*} \int\sec x\,dx\amp = \int{\sec^2 x +\sec x\tan x\over \sec x +\tan x}\,dx\\ \amp = \int{1\over u}\,du=\ln |u|+C\\ \amp = \ln|\sec x +\tan x|+C\text{.} \end{align*}

Now we compute the integral $\ds \int\sec^3 x\,dx\text{.}$

###### Example2.22. Integral of Secant Cubed.

Evaluate $\ds\int\sec^3 x\,dx\text{.}$

Solution
\begin{align*} \sec^3x\amp = {\sec^3x\over2}+{\sec^3x\over2}={\sec^3x\over2}+{(\tan^2x+1)\sec x\over 2}\\ \amp = {\sec^3x\over2}+{\sec x \tan^2 x\over2}+{\sec x\over 2}\\ \amp = {\sec^3x+\sec x \tan^2x\over 2}+{\sec x\over 2}\text{.} \end{align*}

We already know how to integrate $\sec x\text{,}$ so we just need the first quotient. This is “simply” a matter of recognizing the Product Rule differentiation in action:

\begin{equation*} \int \sec^3x+\sec x \tan^2x\,dx=\sec x \tan x\text{.} \end{equation*}

So putting these together we get

\begin{equation*} \int\sec^3x\,dx={\sec x \tan x\over2}+{\ln|\sec x +\tan x|\over2}+C\text{,} \end{equation*}

### Subsection2.2.3Products of Powers of Secant and Tangent

For products of secant and tangent it is best to use the following Guideline.

###### Guideline for Integrating Products of Secant and Tangent.

When evaluating $\ds\int\sec^mx \tan^nx\,dx\text{:}$

1. The power of secant is even ($m$ even):

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

2. Cancel $\sec^2 x$ by the substitution of $dx\text{,}$ and be left with an even number of secants.

3. Use $\sec^2x=1+\tan^2x~(=1+u^2)$ to replace the leftover secants.

2. The power of tangent is odd ($n$ odd):

1. Use $u=\sec x$ and $du=\sec x\tan x\,dx\text{.}$

2. Cancel one $\sec x$ and one $\tan x$ by the substitution of $dx\text{.}$ The number of remaining tangents is even.

3. Use $\tan^2x=\sec^2x-1~(=u^2-1)$ to replace the leftover tangents.

3. $m$ is even or $n$ is odd: Use either $1$ or $2$ (both will work).

4. The power of secant is odd and the power of tangent is even: No guideline. The integrals $\ds\int\sec x\,dx$ and $\ds\int\sec^3 x\,dx$ can usually be looked up, or recalled from memory.

###### Example2.23. Even Power of Secant.

Evaluate $\ds\int\sec^6x\tan^6x\,dx\text{.}$

Solution

Since the power of secant is even, we use $u=\tan x\text{,}$ so that $du=\sec^2x\,dx\text{.}$

\begin{align*} \int \sec^6x\tan^6 x\,dx \amp = \int \sec^6x\,(u^6)\,\ds\frac{du}{\sec^2x} \amp \amp \mbox{Using the substitution} \\ =\amp \int \sec^4x(u^6)\,du \amp\amp \mbox{Cancelling a $\sec^2x$} \\ =\amp \int (\sec^2x)^2(u^6)\,du \amp \amp \mbox{Rewriting $\sec^4x$} \\ =\amp \int (1+\tan^2x)^2(u^6)\,du \amp \amp \mbox{Using $\sec^2x=1+\tan^2x$}\\ =\amp \int (1+u^2)^2(u^6)\,du \amp\amp \mbox{Using the substitution} \end{align*}

To integrate this product the easiest method is expand it into a polynomial and integrate term-by-term.

\begin{align*} \int \sec^6x\tan^6 x\,dx = \amp\int ( u^6+2u^8+u^{10})\,du \amp\amp \mbox{Expanding} \\ =\amp \frac{u^7}{7}+\frac{2u^9}{9}+\frac{u^{11}}{11}+C \amp\amp \mbox{Integrating} \\ = \amp \frac{\tan^7x}{7}+\frac{2\tan^9x}{9}+\frac{\tan^{11}x}{11}+C \amp \amp \mbox{Rewriting in terms of $x$} \end{align*}
###### Example2.24. Odd Power of Tangent.

Evaluate $\ds\int\sec^5x\tan x\,dx\text{.}$

Solution

Since the power of tangent is odd, we use $u=\sec x\text{,}$ so that $du= \sec x\tan x\,dx\text{.}$ Then we have:

\begin{align*} \int \sec^5x\tan x\,dx =\amp \int \sec^5x\tan x\,\ds\frac{du}{\sec x\tan x} \amp\amp \mbox{Substituting $dx$ first} \\ =\amp \int \sec^4x\,du \amp\amp \mbox{Cancelling}\\ =\amp \int u^4\,du \amp\amp \mbox{Using the substitution} \\ =\amp \frac{u^5}{5}+C \amp \amp \mbox{Integrating}\\ = \amp \frac{\sec^5x}{5}+C \amp\amp \mbox{Rewriting in terms of $x$} \end{align*}
###### Example2.25. Odd Power of Secant and Even Power of Tangent.

Evaluate $\ds\int\sec x\tan^2x\,dx\text{.}$

Solution

The Guideline doesn't help us in this scenario. However, since $\ds\tan^2x=\sec^2x-1\text{,}$ we have

\begin{equation*} \begin{split} \int \sec x\tan^2x\,dx = \amp \int \sec x(\sec^2x-1) \,dx \amp \\ = \amp \int (\sec^3x-\sec x)\,dx \\ = \amp \frac{1}{2}\left(\sec x\tan x+\ln|\sec x+\tan x|\right) - \ln|\sec x+\tan x|+C \\ = \amp \frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x+\tan x| - \ln|\sec x+\tan x|+C \\ = \amp \frac{1}{2}\sec x\tan x-\frac{1}{2}\ln|\sec x+\tan x|+C \end{split} \end{equation*}
##### Exercises for Section 2.2.

Evaluate the following indefinite integrals.

1. $\ds\int \sin^2 x\,dx$

$x/2-\sin(2x)/4+C$
Solution

We recall the following identity:

\begin{equation*} \sin^2 x = \frac{1}{2}(1-\cos(2x))\text{.} \end{equation*}

Therefore,

\begin{equation*} \int \sin^2 x \,dx = \frac{1}{2} \int 1-\cos (2x) \,dx\text{.} \end{equation*}

Now let $u=2x$ with $du = 2\,dx\text{:}$

\begin{equation*} \begin{split} \frac{1}{2} \int 1-\cos (2x) \,dx \amp = \frac{1}{4} \int (1-\cos(u))\,du\\ \amp = \frac{1}{4}\left(u - \sin u\right) + C \end{split} \end{equation*}

Hence,

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

$\ds -\cos x+(\cos^3x)/3+C$
Solution

To solve this integral, we will use the substitution $u=\cos x$ with $du = -\sin x\,dx\text{.}$ Then:

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

$3x/8-(\sin 2x)/4+(\sin 4x)/32+C$
Solution

We recall the following identities:

\begin{equation*} \sin^4 x = \frac{1}{4}(1-\cos(2x))^2 \ \text{ and } \ \cos^2 x = \frac{1}{2}\left(1+\cos(2x)\right)\text{.} \end{equation*}

Then the integral becomes

\begin{equation*} \begin{split} \int \sin^4x \,dx \amp = \int \frac{1}{4}(1-\cos(2x))^2 \, dx\\ \amp = \frac{1}{4} \int \left(1-2\cos(2x) + \cos^2(2x)\right)\,dx\\ \amp = \frac{1}{4} \int \left(1-2\cos(2x) + \frac{1}{2}(1+\cos(4x))\right)\,dx\\ \amp = \frac{1}{4}\int\,dx - \frac{1}{2}\int \cos(2x)\,dx + \frac{1}{8}\int 1+\cos(4x)\,dx \\ \amp = \frac{x}{4} - \frac{1}{2}\int \cos(2x)\,dx + \frac{1}{8}\int 1+\cos(4x)\,dx \end{split} \end{equation*}

To solve the second integral, we let $u=2x$ with $du = 2\,dx\text{.}$ For the last integral, we let $t=4x$ with $dt=4\,dx\text{.}$ Then:

\begin{equation*} \begin{split} \int \sin^4x \,dx \amp = \frac{x}{4} - \frac{1}{4}\int \cos(u)\,du + \frac{1}{32}\int 1+\cos(t)\,dt \\ \amp = \frac{x}{4} - \frac{1}{4}\sin u +\frac{t}{32} + \frac{1}{32} \sin(t)\\ \amp = \frac{x}{4} - \frac{1}{4}\sin (2x) +\frac{x}{8} + \frac{1}{32} \sin(4x)\\ \amp = \frac{3x}{8} - \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) \end{split} \end{equation*}
4. $\ds\int \cos^2 x\sin^3 x\,dx$

$\ds (\cos^5 x)/5-(\cos^3x)/3+C$
Solution

Let $=\cos x$ with $du=-\sin x\,dx\text{,}$ and recall the Pythagorean Identity, $\cos^2 x + \sin^2 x = 1\text{.}$ Then,

\begin{equation*} \begin{split} \int u^2 \sin^2x (-du) \amp = -\int u^2 (1-u^2) du \\[1ex] \amp = \int u^4 - u^2 \, du = \frac{1}{5}u^5 - \frac{1}{3}u^3 + C. \end{split} \end{equation*}

Thus,

\begin{equation*} \int \cos^3x \sin^3x \,dx = \frac{1}{5}\cos^5x - \frac{1}{3}\cos^3 x + \hat{C}\text{.} \end{equation*}
5. $\ds\int \cos^3 x\,dx$

$\ds \sin x-(\sin^3x)/3+C$
Solution

First, rewrite the integrand as

\begin{equation*} \cos^3 x = \cos x (1-\sin^2 x)\text{.} \end{equation*}

Now make the substitution,

\begin{equation*} u= \sin x, \ du = \cos x\,dx\text{.} \end{equation*}

Then,

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

$\ds (\sin^3x)/3-(\sin^5x)/5+C$
Solution

Again, we rewrite the integrand as $\cos^3x \sin^2x = (1-\sin^2x)\cos x \sin^2x\text{.}$ Now let

\begin{equation*} u= \sin x, \ du = \cos x\,dx\text{,} \end{equation*}

and solve

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

$\ds -2(\cos x)^{5/2}/5+C$
Solution

Let $u=\cos x$ with $du = -\sin x \,dx.$ Thus,

\begin{equation*} \int \sin x (\cos x)^{3/2} \,dx = \int -u^{3/2} \,du = \frac{2}{5}u^{5/2} + C = \frac{2}{5} \cos^{5/2} + \hat{C}\text{.} \end{equation*}
8. $\ds\int \sec^2 x\csc^2 x\,dx$

$\tan x-\cot x+C$
Solution

Use the identity: \$\sec^2 x = 1+\tan^2 x\text{.}$ Then the integrand becomes

\begin{equation*} \sec^2 x \csc^2 x = (1+\tan^2x)\csc^2 x = \left(1+\frac{\sin^2 x}{\cos^2 x}\right) \frac{1}{\sin^2x} = \frac{1}{\sin^2 x} + \frac{1}{\cos^2 x}\text{.} \end{equation*}

So we have that

\begin{equation*} \int \sec^2 x \csc^2 x\,dx = \int \frac{1}{\cos^2 x} \,dx + \int \frac{1}{\sin^2 x}\, dx\text{.} \end{equation*}

Now recall that

\begin{equation*} \diff{}{x} \tan x = \frac{1}{\cos^2 x}, \text{ and } \diff{}{x} \cot x = -\frac{1}{\sin^2 x}\text{.} \end{equation*}

Therefore,

\begin{equation*} \int \frac{1}{\cos^2 x} \,dx + \int \frac{1}{\sin^2 x}\, dx = \tan x - \cot x + C\text{.} \end{equation*}
9. $\ds\int \tan^3x \sec x\,dx$

$\ds (\sec^3x)/3-\sec x+C$
10. $\ds\int \left(\frac{1}{\csc x}+\frac{1}{\sec x}\right)\,dx$

$\ds -\cos x+\sin x+C$
11. $\ds\int\frac{\cos^2x+\cos x+1}{\cos^3x}\,dx$
$\ds \frac{3}{2}\ln|\sec x+\tan x|+\tan x+\frac{1}{2}\sec x\tan x+C$
12. $\ds\int x\sec^2(x^2)\tan^4(x^2)\,dx$
$\ds \frac{\tan^5(x^2)}{10}+C$
13. $\ds\int x\sec^2(x^2)\tan^4(x^2)\,dx$
$\ds \frac{\tan^5(x^2)}{10}+C$