Previously, we learned that $\int\sin x\ dx=-\cos x+C\quad\int\cos x\ dx=\sin x+C.$ In this section, we will investigate the integrals of other trigonometric functions.

What is $$\int\sec^{2}x\ dx$$?

Because $$(\tan x)’=\sec^{2}x$$, we have

$\bbox[#F2F2F2,5px,border:2px solid black]{\int\sec^{2}x\ dx=\tan x+C.}$ Similarly $\bbox[#F2F2F2,5px,border:2px solid black]{\int\csc^{2}x\ dx=-\cot x+C.}$

It is worth repeating an important integral that we evaluated previously:

Example 1
Find $\int\tan x\ dx.$

Solution 1

Because $$\tan x=\sin x/\cos x$$ and $$(\cos x)’=-\sin x$$, let $$u=\cos x$$. Then $du=-\sin x\ dx,$ and \begin{aligned} \int\tan x\ dx & =\int\frac{\sin x}{\cos x}dx\\ & =\int\frac{-du}{u}\\ & =-\ln|u|+C\\ & =-\ln|\cos x|+C.\end{aligned} Recall that $$\ln A^{B}=B\ln A$$. So \begin{aligned} -\ln|\cos x|+C & =\ln\left(|\cos x|^{-1}\right)+C\\ & =\ln|\sec x|+C.&& {\small(\sec x=\frac{1}{\cos x})}\end{aligned} Therefore, the result can also be written as $\int\tan x\ dx=\ln|\sec x|+C.$

$\bbox[#F2F2F2,5px,border:2px solid black]{\int\tan x\ dx=\ln|\sec x|+C=-\ln|\cos x|+C.}$

Example 2
Find $\int\cot x\ dx.$

Solution 2

Because $$\cot x=\cos x/\sin x$$, if we let $$u=\sin x$$, then $du=\cos x\ dx$ and \begin{aligned} \int\cot x\ dx & =\int\frac{\cos x}{\sin x}dx\\ & =\int\frac{du}{u}\\ & =\ln|u|+C\\ & =\ln|\sin x|+C.\end{aligned}

$\bbox[#F2F2F2,5px,border:2px solid black]{\int\cot x\ dx=\ln|\sin x|+C.}$

Example 3

Find $\int\sec x\tan x\ dx.$

Solution 3

Because $\frac{d}{dx}\sec x=\sec x\tan x$ we already know that the result of this integral is $$\sec x+C$$. However, even if we did not know the result, we could use the integration by substitution to derive the result: \begin{aligned} \int\sec x\tan x\ dx & =\int\frac{1}{\cos x}\frac{\sin x}{\cos x}dx\\ & =\int\frac{\sin x}{\cos^{2}x}dx\end{aligned} Let $$u=\cos x$$, then $du=-\sin x\ dx.$ So \begin{aligned} \int\frac{\sin x}{\cos^{2}x}dx & =\int-\frac{du}{u^{2}}\\ & =-\int u^{-2}du\\ & =-(-u^{-1})+C\\ & =\frac{1}{u}+C\\ & =\frac{1}{\cos x}+C && \small{(u=\cos x)}\\ & =\sec x+C\end{aligned}

Example 4

Find $${\displaystyle \int e^{x}\sec(e^{x})\tan(e^{x})\ dx}$$.

Solution 4

Let $$u=e^{x}$$. Then $du=e^{x}\ dx$ and \begin{aligned} \int\sec(\underbrace{e^{x}}_{u})\tan(\underbrace{e^{x}}_{u})\underbrace{e^{x}dx}_{du} & =\int\sec u\tan u\ du\\ & =\sec u+C\\ & =\sec e^{x}+C.\end{aligned}

Example 5

Find $${\displaystyle \int}\sec x\ dx$$

Solution 5

There are different ways to evaluate this integral; each involves a trick. The easiest way is to multiply and divide the integrand by $$\sec x+\tan x$$: \begin{aligned} \int\sec x\ dx & =\int\sec x\frac{\sec x+\tan x}{\sec x+\tan x}dx\\ & =\int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}dx.\end{aligned} Let $$u=\sec x+\tan x$$. Then $du=\left(\sec x\tan x+\sec^{2}x\right)dx.$ So \begin{aligned} \int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}dx & =\int\frac{du}{u}\\ & =\ln|u|+C\\ & =\ln|\sec x+\tan x|+C.\end{aligned}

$\bbox[#F2F2F2,5px,border:2px solid black]{\int\sec x\ dx=\ln|\sec x+\tan x|+C.}$

Example 6

Find $${\displaystyle \int\sin^{2}x\ dx}$$.

Solution 6

Because $$\sin^{2}x=(1-\cos2x)/2$$, we have \begin{aligned} \int\sin^{2}x\ dx & =\frac{1}{2}\int(1-\cos2x)dx\\ & =\frac{1}{2}x-\frac{1}{4}\sin2x+C.\end{aligned} To find the integral of $$\cos2x$$, we let $$u=2x$$.

Example 7

Find $${\displaystyle \int\cos^{2}x\ dx}$$.

Solution 7

Because $$\cos^{2}x=(1+\cos2x)/2$$, we have \begin{aligned} \int\cos^{2}x\ dx & =\frac{1}{2}\int(1+\cos2x)dx\\ & =\frac{1}{2}x+\frac{1}{4}\sin2x+C.\end{aligned}

In many applied problems such as mechanical vibrations and electromagnetic waves, we encounter the following trigonometric integrals $\int\sin ax\cos bx\ dx,\quad\int\sin ax\sin bx\ dx,\quad\int\cos ax\cos bx.$ To evaluate these integrals, we use the following product-to-sum identities (also see here)

$\bbox[#F2F2F2,5px,border:2px solid black]{\sin ax\ \cos bx=\frac{1}{2}\left[\sin\,(a-b)x+\sin\,(a+b)x\right]}$

$\bbox[#F2F2F2,5px,border:2px solid black]{\sin ax\ \sin bx=\frac{1}{2}\left[\cos\,(a-b)x-\cos\,(a+b)x\right]}$

$\bbox[#F2F2F2,5px,border:2px solid black]{\cos ax\ \cos bx=\frac{1}{2}\left[\cos\,(a-b)x+\cos\,(a+b)x\right]}$

Example 8

Find $${\displaystyle \int\sin3x\,\cos5x\,dx}$$.

Solution 8

\begin{aligned} \int\sin3x\,\cos5x\,dx & =\frac{1}{2}\int\left[\sin(-2x)+\sin8x\right]dx\\ & =\frac{1}{2}\int\left[\sin8x-\sin2x\right]\ dx\\ & =\frac{1}{2}\left(-\frac{1}{8}\cos8x+\frac{1}{2}\cos2x\right)+C\\ & =\frac{1}{4}\cos2x-\frac{1}{16}\cos8x+C.\end{aligned}

### Finding $${\displaystyle \int\sin^{m}x\cos^{n}dx}$$ when either $$m$$ or $$n$$ is a positive odd integer.

When $$\sin x$$ has the odd exponent, we first reduce the integral to the form $\int(\text{terms involving only }\cos x)\sin xdx,$ and when $$\cos x$$ has the odd exponent, we reduce the integral to the form $\int(\text{terms involving only }\sin x)\cos xdx.$ Then we can perform the integration by means of $\int u^{n}du=\frac{u^{n+1}}{n+1}+C.$ The following examples will illustrate how to use this.

Example 9

Find $${\displaystyle \int\sin^{2}x\cos^{5}x\,dx}$$

Solution 9

Here the exponent of $$\cos x$$ is odd. Thus \begin{aligned} {\displaystyle \int\sin^{2}x\cos^{5}xdx} & =\int\sin^{2}x\cos^{4}x\cos xdx\\ & =\int\sin^{2}x\,(1-\sin^{2}x)^{2}\cos xdx\\ & =\int(\sin^{2}x-2\sin^{4}x+\sin^{6}x)\cos xdx\\ & =\int\sin^{2}x\cos xdx-2\int\sin^{4}x\cos xdx+\int\sin^{6}x\cos xdx\end{aligned} Now let $$\sin x=u\Rightarrow\cos xdx=du$$. Thus $\int\sin^{2}x\cos^{5}xdx=\frac{\sin^{3}x}{3}-2\frac{\sin^{5}x}{5}+\frac{\sin^{7}x}{7}+C.$

Example 10

Find $${\displaystyle \int\cos^{3}xdx}$$

Solution 10

\begin{aligned} \int\cos^{3}xdx & =\int\cos^{2}x\cos xdx\\ & =\int(1-\sin^{2}x)\cos xdx\\ & =\int\cos xdx-\int\sin^{2}x\cos xdx\\ & =\sin x-\frac{\sin^{2}x}{3}+C.\end{aligned}

Example 11

Find $${\displaystyle \int\frac{\cos^{5}x}{\sqrt{\sin x}}dx}$$.

Solution 11

Here $$m=-1/2$$ and $$n=5$$. We separate a factor of $$\cos x$$ from $$\cos^{5}x$$ and express the remaining factor $$\cos^{4}x$$ in terms of $$\sin x$$, That is, \begin{aligned} \int\frac{\cos^{5}x}{\sqrt{\sin x}}dx & =\int\frac{\cos^{4}x}{\sqrt{\sin x}}\cos x\ dx\\ & =\int\frac{(1-\sin^{2}x)^{2}}{\sqrt{\sin x}}\cos x\ dx.\end{aligned} Now let $$u=\sin x$$. Then $$du=\cos x\ dx$$ and \begin{aligned} \int\frac{(1-\sin^{2}x)^{2}}{\sqrt{\sin x}}\cos x\ dx & =\int\frac{(1-u^{2})^{2}}{\sqrt{u}}du\\ & =\int\frac{1-2u^{2}+u^{4}}{\sqrt{u}}du\\ & =\int\left(u^{-1/2}-2u^{3/2}+u^{7/2}\right)du\\ & =2u^{1/2}-\frac{4}{5}u^{5/2}+\frac{2}{9}u^{9/2}+C\\ & =2\sqrt{\sin x}-\frac{4}{5}\sqrt{\sin^{5}x}+\frac{2}{9}\sqrt{\sin^{9}x}+C.\end{aligned}