# Integration of Some Classes of Trigonometric Functions

• In this section we consider $$8$$ classes of integrals with trigonometric functions. Special transformations and substitutions used for each of these classes allow us to obtain exact solutions for these integrals.

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

To find integrals of this type, use the following trigonometric identities:

• $${\cos ax\cos bx }$$ $$= {{\large\frac{1}{2}\normalsize} \left[ {\cos \left( {ax + bx} \right) }\right.}$$ $$+\, {\left.{ \cos \left( {ax – bx} \right)} \right]}$$
• $${\sin ax\cos bx }$$ $$= {{\large\frac{1}{2}\normalsize} \left[ {\sin \left( {ax + bx} \right) }\right.}$$ $$+\, {\left.{ \sin \left( {ax – bx} \right)} \right]}$$
• $${\sin ax\sin bx }$$ $$= {-{\large\frac{1}{2}\normalsize} \left[ {\cos \left( {ax + bx} \right) }\right.}$$ $$-\, {\left.{ \cos \left( {ax – bx} \right)} \right]}$$

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

It’s assumed here and below that $$m$$ and $$n$$ are positive integers. To find an integral of this form, use the following substitutions:

1. If the power $$n$$ of the cosine is odd (the power $$m$$ of the sine can be arbitrary), then the substitution $$u = \sin x$$ is used.
2. If the power $$m$$ of the sine is odd, then the substitution $$u = \cos x$$ is used.
3. If both powers $$m$$ and $$n$$ are even, then first use the double angle formulas $${\sin 2x = 2\sin x\cos x,}$$ $${\cos 2x }$$ $$={ {\cos^2}x – {\sin ^2}x }$$ $$= {1 – 2\,{\sin ^2}x }$$ $$= {2\,{\cos ^2}x – 1}$$ to reduce the power of the sine or cosine in the integrand. Then, if necessary, apply the rules $$1$$ or $$2.$$

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

The power of the integrand can be reduced by 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 $${\large\int\normalsize} {{{\cot }^n}xdx}$$

We can reduce the power of the integrand using the trigonometric identity $$1 + {\cot ^n}x = {\csc ^n}x$$ 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 $${\large\int\normalsize} {{\sec^n}xdx}$$

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 $${\large\int\normalsize} {{\csc^n}xdx}$$

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 $${\large\int\normalsize} {{\tan^m}x\,{\sec^n}xdx}$$

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 $${\large\int\normalsize} {{\cot^m}x\,{\csc^n}xdx}$$

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 a problem to see the solution.

### Example 1

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

### Example 2

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

### Example 3

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

### Example 4

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

### Example 5

Find the integral $${\large\int\normalsize} {{{\sin }^3}x\sqrt {\cos x} dx}.$$

### Example 6

Evaluate the integral $${\large\int\normalsize} {\sin {\large\frac{x}{4}\normalsize} \cos {\large\frac{x}{3}\normalsize} dx}.$$

### Example 7

Evaluate the integral $${\large\int\normalsize} {{{\tan }^4}xdx}.$$

### Example 8

Calculate the integral $${\large\int\normalsize} {{{\cot }^5}xdx}.$$

### Example 9

Calculate the integral $${\large\int\normalsize} {{{\sec }^3}xdx}.$$

### Example 10

Evaluate the integral $${\large\int\normalsize} {{{\csc }^4}xdx}.$$

### Example 11

Compute $${\large\int\normalsize} {{\tan^3}x\,{{\sec }^2}xdx}.$$

### Example 12

Compute $${\large\int\normalsize} {{\tan^2}x\sec xdx}.$$

### Example 1.

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

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 $${\large\int\normalsize} {{\cos^5}xdx}.$$

Solution.

Making the substitution $$u = \sin x,$$ $$du = \cos xdx$$ and using the identity $${\cos ^2}x = 1 – {\sin ^2}x,$$ we obtain

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

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Problems 1-2
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Problems 3-12