# Calculus/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
    $I$ $={\frac {1}{8}}\left(\int 1dx-\int \cos ^{2}(2x)dx+\int \cos(2x)dx-\int \cos ^{3}(2x)dx\right)$ $={\frac {1}{8}}\left(x-{\frac {1}{2}}\int {\Big1+\cos(4x){\Big )}dx+{\frac {\sin(2x)}{2}}-\int \cos ^{2}(2x)\cos(2x)dx\right)$ TODO: CORRECT FORMULA$={\frac {1}{164}}\left(x+\sin(2x)+\int \cos(4x)dx-2\int {\Big1-\sin ^{2}(2x){\Big )}\cos(2x)dx\right)$
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 $\int \tan ^{m}(x)\sec ^{n}(x)dx$ .
   * If $n$ is even and $n\geq 2$ then substitute $u=tan(x)$ and use the identity $\sec ^{2}(x)=1+\tan ^{2}(x)$ .
   * If $n$ and $m$ are both odd then substitute $u=\sec(x)$ and use the identity $\tan ^{2}(x)=\sec ^{2}(x)-1$ .
   * 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
    $\int \sin(x)\sin(2x)dx$ $={\frac {1}{2}}\int {\Big\cos(x)-\cos(3x){\Big )}dx$ $={\frac {\sin(x)}{2}}-{\frac {\sin(3x)}{6}}+C$
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