# Calculus/Integration techniques/Trigonometric Integrals

 ← Integration techniques/Trigonometric Substitution Calculus Integration techniques/Partial Fraction Decomposition → Integration techniques/Trigonometric Integrals

When the integrand is primarily or exclusively based on trigonometric functions, the following techniques are useful.

### Powers of Sine and Cosine

We will give a general method to solve generally integrands of the form $\cos ^{m}(x)\cdot \sin ^{n}(x)$  . First let us work through an example.

$\int \cos ^{3}(x)\sin ^{2}(x)dx$

Notice that the integrand contains an odd power of cos. So rewrite it as

$\int \cos ^{2}(x)\sin ^{2}(x)\cos(x)dx$

We can solve this by making the substitution $u=\sin(x)$  so $du=\cos(x)dx$  . Then we can write the whole integrand in terms of $u$  by using the identity

$\cos ^{2}(x)=1-\sin ^{2}(x)=1-u^{2}$  .

So

 $\int \cos ^{3}(x)\sin ^{2}(x)dx$ $=\int \cos ^{2}(x)\sin ^{2}(x)\cos(x)dx$ $=\int (1-u^{2})u^{2}du$ $=\int u^{2}du-\int u^{4}du$ $={\frac {u^{3}}{3}}+{\frac {u^{5}}{5}}+C$ $={\frac {\sin ^{3}(x)}{3}}-{\frac {\sin ^{5}(x)}{5}}+C$ This method works whenever there is an odd power of sine or cosine.

To evaluate $\int \cos ^{m}(x)\sin ^{n}(x)dx$  when either $m$  or $n$  is odd.

• If $m$  is odd substitute $u=\sin(x)$  and use the identity $\cos ^{2}(x)=1-\sin ^{2}(x)=1-u^{2}$  .
• If $n$  is odd substitute $u=\cos(x)$  and use the identity $\sin ^{2}(x)=1-\cos ^{2}(x)=1-u^{2}$  .

#### Example

Find $\int \limits _{0}^{\frac {\pi }{2}}\cos ^{40}(x)\sin ^{3}(x)dx$  .

As there is an odd power of $\sin$  we let $u=\cos(x)$  so $du=-\sin(x)dx$  . Notice that when $x=0$  we have $u=\cos(0)=1$  and when $x={\frac {\pi }{2}}$  we have $u=\cos({\tfrac {\pi }{2}})=0$  .

 $\int \limits _{0}^{\frac {\pi }{2}}\cos ^{40}(x)\sin ^{3}(x)dx$ $=\int \limits _{0}^{\frac {\pi }{2}}\cos ^{40}(x)\sin ^{2}(x)\sin(x)dx$ $=-\int \limits _{1}^{0}u^{40}(1-u^{2})du$ $=\int \limits _{0}^{1}u^{40}(2-u^{2})du$ $=\int \limits _{0}^{5}(u^{40}-u^{42})du$ $=\left({\frac {u^{41}}{41}}-{\frac {u^{43}}{43}}\right){\Bigg |}_{0}^{1}$ $={\frac {1}{41}}-{\frac {1}{43}}$ When both $m$  and $n$  are even, things get a little more complicated.

To evaluate $\int \cos ^{m}(x)\sin ^{n}(x)dx$  when both $m$  and $n$  are even.

Use the identities $\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}$  and $\cos ^{2}(x)={\frac {1+\cos(2x)}{2}}$  .

#### Example

Find $\int \sin ^{2}(x)\cos ^{4}(x)dx$  .

As $\sin ^{2}(x)={\frac {1-\cos(2x)}{2}}$  and $\cos ^{2}(x)={\frac {1+\cos(2x)}{2}}$  we have

$\int \sin ^{2}(x)\cos ^{4}(x)dx=\int \left({\frac {1-\cos(2x)}{2}}\right)\left({\frac {1+\cos(2x)}{2}}\right)^{2}dx$

and expanding, the integrand becomes

${\frac {1}{8}}\int \left(1-\cos ^{2}(2x)+\cos(2x)-\cos ^{3}(2x)\right)dx$

Using the multiple angle identities

 $I$ $={\frac {1}{8}}\left(\int 1dx-\int \cos ^{2}(2x)dx+\int \cos(2x)dx-\int \cos ^{3}(2x)dx\right)$ $={\frac {1}{8}}\left(x-{\frac {1}{2}}\int {\Big (}1+\cos(4x){\Big )}dx+{\frac {\sin(2x)}{2}}-\int \cos ^{2}(2x)\cos(2x)dx\right)$ TODO: CORRECT FORMULA$={\frac {1}{164}}\left(x+\sin(2x)+\int \cos(4x)dx-2\int {\Big (}1-\sin ^{2}(2x){\Big )}\cos(2x)dx\right)$ then we obtain on evaluating

$I={\frac {x}{16}}-{\frac {\sin(4x)}{64}}+{\frac {\sin ^{3}(2x)}{48}}+C$

### Powers of Tan and Secant

To evaluate $\int \tan ^{m}(x)\sec ^{n}(x)dx$  .

1. If $n$  is even and $n\geq 2$  then substitute $u=tan(x)$  and use the identity $\sec ^{2}(x)=1+\tan ^{2}(x)$  .
2. If $n$  and $m$  are both odd then substitute $u=\sec(x)$  and use the identity $\tan ^{2}(x)=\sec ^{2}(x)-1$  .
3. If $n$  is odd and $m$  is even then use the identity $\tan ^{2}(x)=\sec ^{2}(x)-1$  and apply a reduction formula to integrate $\sec ^{j}(x)dx$  , using the examples below to integrate when $j=1,2$  .

#### Example 1

Find $\int \sec ^{2}(x)dx$  .

There is an even power of $\sec(x)$  . Substituting $u=\tan(x)$  gives $du=\sec ^{2}(x)dx$  so

$\int \sec ^{2}(x)dx=\int du=u+C=\tan(x)+C.$

#### Example 2

Find $\int \tan(x)dx$  .

Let $u=\cos(x)$  so $du=-\sin(x)dx$  . Then

 $\int \tan(x)dx$ $=\int {\frac {\sin(x)}{\cos(x)}}dx$ $=\int -{\frac {du}{u}}$ $=-\ln |u|+C$ $=-\ln {\Big |}\cos(x){\Big |}+C$ $=\ln {\Big |}\sec(x){\Big |}+C$ #### Example 3

Find $\int \sec(x)dx$  .

The trick to do this is to multiply and divide by the same thing like this:

 $\int \sec(x)dx$ $=\int \sec(x){\frac {\sec(x)+\tan(x)}{\sec(x)+\tan(x)}}dx$ $=\int {\frac {\sec ^{2}(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}}dx$ Making the substitution $u=\sec(x)+\tan(x)$  so $du=\sec(x)\tan(x)+\sec ^{2}(x)dx$  ,

 $\int \sec(x)dx$ $=\int {\frac {du}{u}}$ $=\ln |u|+C$ $\ln {\Big |}\sec(x)+\tan(x){\Big |}+C$ ### More trigonometric combinations

For the integrals $\int \sin(nx)\cos(mx)dx$  or $\int \sin(nx)\sin(mx)dx$  or $\int \cos(nx)\cos(mx)dx$  use the identities

• $\sin(a)\cos(b)={\frac {\sin(a+b)+\sin(a-b)}{2}}$
• $\sin(a)\sin(b)={\frac {\cos(a-b)-\cos(a+b)}{2}}$
• $\cos(a)\cos(b)={\frac {\cos(a-b)+\cos(a+b)}{2}}$

#### Example 1

Find $\int \sin(3x)\cos(5x)dx$  .

We can use the fact that $\sin(a)\cos(b)={\frac {\sin(a+b)+\sin(a-b)}{2}}$  , so

$\sin(3x)\cos(5x)={\frac {\sin(8x)+\sin(-2x)}{2}}$

Now use the oddness property of $\sin(x)$  to simplify

$\sin(3x)\cos(5x)={\frac {\sin(8x)-\sin(2x)}{2}}$

And now we can integrate

 $\int \sin(3x)\cos(5x)dx$ $=\int {\Big (}{\frac {\sin(8x)-\sin(2x)}{2}}{\Big )}dx$ $={\frac {\cos(2x)}{4}}-{\frac {\cos(8x)}{16}}+C$ #### Example 2

Find:$\int \sin(x)\sin(2x)dx$  .

Using the identities

$\sin(x)\sin(2x)={\frac {\cos(-x)-\cos(3x)}{2}}={\frac {\cos(x)-\cos(3x)}{2}}$

Then

 $\int \sin(x)\sin(2x)dx$ $={\frac {1}{2}}\int {\Big (}\cos(x)-\cos(3x){\Big )}dx$ $={\frac {\sin(x)}{2}}-{\frac {\sin(3x)}{6}}+C$ ← Integration techniques/Trigonometric Substitution Calculus Integration techniques/Partial Fraction Decomposition → Integration techniques/Trigonometric Integrals