# Trigonometric Integrals

## Trigonometry # Trigonometric Integrals

In this topic, we will study how to integrate certain combinations involving products and powers of trigonometric functions.

We consider $$8$$ cases.

## 1. Integrals of the form $$\int {\cos ax\cos bxdx} ,$$ $$\int {\sin ax\cos bxdx} ,$$ $$\int {\sin ax\sin bxdx}$$

To evaluate integrals of products of sine and cosine with different arguments, we apply the identities

$\cos ax\cos bx = \frac{1}{2}\left[ {\cos \left( {ax + bx} \right) + \cos \left( {ax - bx} \right)} \right];$
$\sin ax\cos bx = \frac{1}{2}\left[ {\sin \left( {ax + bx} \right) + \sin \left( {ax - bx} \right)} \right];$
$\sin ax\sin bx = - \frac{1}{2}\left[ {\cos \left( {ax + bx} \right) - \cos \left( {ax - bx} \right)} \right].$

## 2. Integrals of the form $$\int {{\sin^m}x\,{\cos^n}xdx}$$

We assume here that the powers $$m$$ and $$n$$ are non-negative integers.

To find an integral of this form, use the following substitutions:

1. If $$m$$ (the power of sine) is odd, we use the $$u-$$substitution
$u = \cos x,\;\; du = - \sin xdx$

and the identity

${\sin ^2}x + {\cos ^2}x = 1$

to express the remaining even power of sine in $$u-$$terms.

2. If $$n$$ (the power of cosine) is odd, we use the $$u-$$substitution
$u = \sin x,\;\; du = \cos xdx$

and the identity

${\sin ^2}x + {\cos ^2}x = 1$

to express the remaining even power of cosine in $$u-$$terms.

3. If both powers $$m$$ and $$n$$ are even, we reduce the powers using the half-angle formulas
${\sin ^2}x = \frac{{1 - \cos 2x}}{2},\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2}.$

The integrals of type $$\int {{{\sin }^n}xdx}$$ and $$\int {{{\cos }^n}xdx}$$ can be evaluated by reduction formulas

$\int {{{\sin }^n}xdx} = - \frac{{{{\sin }^{n - 1}}x\cos x}}{n} + \frac{{n - 1}}{n}\int {{{\sin }^{n - 2}}xdx} ,$
$\int {{{\cos }^n}xdx} = \frac{{{{\cos }^{n - 1}}x\sin x}}{n} + \frac{{n - 1}}{n}\int {{{\cos }^{n - 2}}xdx} .$

## 3. Integrals of the form $$\int {{\tan^n}xdx}$$

The power of the integrand can be reduced using the trigonometric identity

$1 + {\tan ^2}x = {\sec ^2}x$

and the reduction formula

$\int {{\tan^n}xdx} = \int {{{\tan }^{n - 2}}x\,{{\tan }^2}xdx} = \int {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)dx} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}xdx} .$

## 4. Integrals of the form $$\int {{\cot^n}xdx}$$

The power of the integrand can be reduced using the trigonometric identity

$1 + {\cot^2}x = {\csc^2}x$

and the reduction formula

$\int {{{\cot }^n}xdx} = \int {{{\cot }^{n - 2}}x\,{{\cot }^2}xdx} = \int {{{\cot }^{n - 2}}x\left( {{{\csc }^2}x - 1} \right)dx} = - \frac{{{{\cot }^{n - 1}}x}}{{n - 1}} - \int {{{\cot }^{n - 2}}xdx} .$

## 5. Integrals of the form $$\int {{\sec^n}xdx}$$

This type of integrals can be simplified with help of the reduction formula:

$\int {{\sec^n}xdx} = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{\sec^{n - 2}}xdx} .$

## 6. Integrals of the form $$\int {{\csc^n}xdx}$$

Similarly to the previous examples, this type of integrals can be simplified by the formula

$\int {{\csc^n}xdx} = - \frac{{{\csc^{n - 2}}x \cot x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{\csc^{n - 2}}xdx} .$

## 7. Integrals of the form $$\int {{\tan^m}x\,{\sec^n}xdx}$$

1. If the power of the secant $$n$$ is even, then using the identity
$1 + {\tan ^2}x = {\sec ^2}x$

the secant function is expressed as the tangent function. The factor $${\sec ^2}x$$ is separated and used for transformation of the differential. As a result, the entire integral (including differential) is expressed in terms of the function $$\tan x.$$

2. If both the powers $$n$$ and $$m$$ are odd, then the factor $$\sec x \tan x,$$ which is necessary to transform the differential, is separated. Then the entire integral is expressed in terms of $$\sec x.$$
3. If the power of the secant $$n$$ is odd, and the power of the tangent $$m$$ is even, then the tangent is expressed in terms of the secant using the identity
$1 + {\tan ^2}x = {\sec ^2}x.$

After this substitution, you can calculate the integrals of the secant.

## 8. Integrals of the form $$\int {{\cot^m}x\,{\csc^n}xdx}$$

1. If the power of the cosecant $$n$$ is even, then using the identity
$1 + {\cot^2}x = {\csc ^2}x$

the cosecant function is expressed as the cotangent function. The factor $${\csc^2}x$$ is separated and used for transformation of the differential. As a result, the integrand and differential are expressed in terms of $$\cot x.$$

2. If both the powers $$n$$ and $$m$$ are odd, then the factor $$\cot x \csc x,$$ which is necessary to transform the differential, is separated. Then the integral is expressed in terms of $$\csc x.$$
3. If the power of the cosecant $$n$$ is odd, and the power of the cotangent $$m$$ is even, then the cotangent is expressed in terms of the cosecant using the identity
$1 + {\cot^2}x = {\csc ^2}x.$

After this substitution, you can find the integrals of the cosecant.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Calculate the integral $\int {{\sin^3}xdx}.$

### Example 2

Evaluate the integral $\int {{\cos^5}xdx}.$

### Example 3

Find the integral $\int {{\sin^6}xdx}.$

### Example 4

Find the integral $\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.$

### Example 5

Calculate the integral $\int {{{\sin }^2}x\,{{\cos }^4}xdx}.$

### Example 6

Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^4}xdx}.$

### Example 7

Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^5}xdx}.$

### Example 8

Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^3}xdx}.$

### Example 1.

Calculate the integral $\int {{\sin^3}xdx}.$

Solution.

Let $$u = \cos x,$$ $$du = -\sin xdx.$$ Then

$\int {{\sin^3}xdx} = \int {{\sin^2}x\sin xdx} = \int {\left( {1 - {\cos^2}x} \right)\sin xdx} = - \int {\left( {1 - {u^2}} \right)du} = \int {\left( {{u^2} - 1} \right)du} = \frac{{{u^3}}}{3} - u + C = \frac{{{{\cos }^3}x}}{3} - \cos x + C.$

### Example 2.

Evaluate the integral $\int {{\cos^5}xdx}.$

Solution.

Making the substitution $$u = \sin x,$$ $$du = \cos xdx$$ and using the identity

${\cos ^2}x = 1 - {\sin ^2}x,$

we obtain:

$\int {{\cos^5}xdx} = \int {{{\left( {{\cos^2}x} \right)}^2}\cos xdx} = \int {{{\left( {1 - {{\sin }^2}x} \right)}^2}\cos x dx} = \int {{{\left( {1 - {u^2}} \right)}^2}du} = \int {\left( {1 - 2{u^2} + {u^4}} \right)du} = u - \frac{{2{u^3}}}{3} + \frac{{{u^5}}}{5} + C = \sin x - \frac{{2{{\sin }^3}x}}{3} + \frac{{{{\sin }^5}x}}{5} + C.$

### Example 3.

Find the integral $\int {{\sin^6}xdx}.$

Solution.

Using the identities

${\sin ^2}x = \frac{{1 - \cos 2x}}{2}\;\;\text{and}\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2},$

we can write:

$I = \int {{\sin^6}xdx} = \int {{{\left( {{\sin^2}x} \right)}^3}dx} = \frac{1}{8}\int {{{\left( {1 - \cos 2x} \right)}^3}dx} = \frac{1}{8}\int {\left( {1 - 3\cos 2x + 3\,{{\cos }^2}2x - {{\cos }^3}2x} \right)dx} = \frac{x}{8} - \frac{3}{8} \cdot \frac{{\sin 2x}}{2} + \frac{3}{8}\int {{\cos^2}2xdx} - \frac{3}{8}\int {{\cos^3}2xdx}.$

Calculate the integrals in the latter expression.

$\int {{\cos^2}2xdx} = \int {\frac{{1 + \cos 4x}}{2}dx} = \frac{1}{2}\int {\left( {1 + \cos 4x} \right)dx} = \frac{1}{2}\left( {x + \frac{{\sin 4x}}{4}} \right) = \frac{x}{2} + \frac{{\sin 4x}}{8}.$

To find the integral $$\int {{\cos^3}2xdx},$$ we make the substitution $$u = \sin 2x,$$ $$du =$$ $$2\cos 2xdx.$$ Then

$u = \sin 2x,\;du = 2\cos 2xdx.$

Then

$\int {{\cos^3}2xdx} = \frac{1}{2}\int {2{{\cos }^2}2x\cos 2xdx} = \frac{1}{2}\int {2\left( {1 - {{\sin }^2}2x} \right)\cos 2xdx} = \frac{1}{2}\int {\left( {1 - {u^2}} \right)du} = \frac{u}{2} - \frac{{{u^3}}}{6} = \frac{{\sin 2x}}{2} - \frac{{{{\sin }^3}2x}}{6}.$

Hence, the initial integral is

$I = \frac{x}{8} - \frac{{3\sin 2x}}{{16}} + \frac{3}{8}\left( {\frac{x}{2} + \frac{{\sin 4x}}{8}} \right) - \frac{1}{8}\left( {\frac{{\sin 2x}}{2} - \frac{{{{\sin }^3}2x}}{6}} \right) + C = \frac{{5x}}{{16}} - \frac{{\sin 2x}}{4} + \frac{{3\sin 4x}}{{64}} + \frac{{{{\sin }^3}2x}}{{48}} + C.$

### Example 4.

Find the integral $\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.$

Solution.

The power of cosine is odd, so we make the substitution

$u = \sin x,\;\;du = \cos xdx.$

We rewrite the integral in terms of $$\sin x$$ to obtain:

$\int {{{\sin }^2}x\,{{\cos }^3}xdx} = \int {{{\sin }^2}x\,{{{\cos }^2}x}\cos xdx} = \int {{{\sin }^2}x{\left( {1 - {{\sin }^2}x} \right)}\cos xdx} = \int {{u^2}\left( {1 - {u^2}} \right)du} = \int {\left( {{u^2} - {u^4}} \right)du} = \frac{{{u^3}}}{3} - \frac{{{u^5}}}{5} + C = \frac{{{{\sin }^3}x}}{3} - \frac{{{{\sin }^5}x}}{5} + C.$

### Example 5.

Calculate the integral $\int {{{\sin }^2}x\,{{\cos }^4}xdx}.$

Solution.

We can write:

$I = \int {{{\sin }^2}x\,{{\cos }^4}xdx} = \int {{{\left( {\sin x\cos x} \right)}^2}{{\cos }^2}xdx} .$

We convert the integrand using the identities

$\sin x\cos x = \frac{{\sin 2x}}{2},\;\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2},\;\;\;{\sin ^2}x = \frac{{1 - \cos 2x}}{2}.$

This yields

$I = \int {{{\left( {\frac{{\sin 2x}}{2}} \right)}^2}\frac{{1 + \cos 2x}}{2}dx} = \frac{1}{8}\int {{{\sin }^2}2x\left( {1 + \cos 2x} \right)dx} = \frac{1}{8}\int {{{\sin }^2}2xdx} + \frac{1}{8}\int {{{\sin }^2}2x\cos 2xdx} = \frac{1}{8}\int {\frac{{1 - \cos 4x}}{2}dx} + \frac{1}{{16}}\int {2{{\sin }^2}2x\cos 2xdx} = \frac{1}{{16}}\int {\left( {1 - \cos 4x} \right)dx} + \frac{1}{{16}}\int {{{\sin }^2}2x\,d\left( {\sin 2x} \right)} = \frac{1}{{16}}\left( {x - \frac{{\sin 4x}}{4}} \right) + \frac{1}{{16}} \cdot \frac{{{{\sin }^3}2x}}{3} + C = \frac{x}{{16}} - \frac{{\sin 4x}}{{64}} + \frac{{{{\sin }^3}2x}}{{48}} + C.$

### Example 6.

Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^4}xdx}.$

Solution.

As the power of sine is odd, we use the substitution

$u = \cos x,\;\;du = - \sin xdx.$

The integral is written as

$I = \int {{{\sin }^3}x\,{{\cos }^4}xdx} = \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} .$

By the Pythagorean identity,

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

Hence

$I = \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^4}x\sin xdx} = - \int {\left( {1 - {u^2}} \right){u^4}du} = \int {\left( {{u^6} - {u^4}} \right)du} = \frac{{{u^7}}}{7} - \frac{{{u^5}}}{5} + C = \frac{{{{\cos }^7}x}}{7} - \frac{{{{\cos }^5}x}}{5} + C.$

### Example 7.

Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^5}xdx}.$

Solution.

We see that both powers are odd, so we can substitute either $$u = \sin x$$ or $$u = \cos x.$$ Choosing the least exponent, we have

$u = \cos x,\;\;du = - \sin xdx.$

The integral takes the form

$I = \int {{{\sin }^3}x\,{{\cos }^5}xdx} = \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} .$

Using the Pythagorean identity

${\sin ^2}x = 1 - {\cos ^2}x,$

we can write

$I = \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^5}x\sin xdx} = - \int {\left( {1 - {u^2}} \right){u^5}du} = \int {\left( {{u^7} - {u^5}} \right)du} = \frac{{{u^8}}}{8} - \frac{{{u^6}}}{6} + C = \frac{{{{\cos }^8}x}}{8} - \frac{{{{\cos }^6}x}}{6} + C.$

### Example 8.

Evaluate the integral $\int {{{\sin }^3}x\,{{\cos }^3}xdx}.$

Solution.

The powers of both sine and cosine are odd. Hence we can use the substitution $$u = \sin x$$ or $$u = \cos x.$$ Let's apply the substitution $$u = \sin x.$$ Then $$du = \cos x dx,$$ and the integral becomes

$I = \int {{{\sin }^3}x\,{{\cos }^3}xdx} = \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} .$

By the Pythagorean identity,

${\sin ^2}x = 1 - {\cos ^2}x,$

so we obtain

$I = \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} = \int {{{\sin }^3}x\left( {1 - {{\sin }^2}x} \right)\cos xdx} = \int {{u^3}\left( {1 - {u^2}} \right)du} = \int {\left( {{u^3} - {u^5}} \right)du} = \frac{{{u^4}}}{4} - \frac{{{u^6}}}{6} + C = \frac{{{{\sin }^4}x}}{4} - \frac{{{{\sin }^6}x}}{6} + C.$