Calculus

Integration of Functions

Integration of Some Classes of Trigonometric Functions

Page 1
Problems 1-2
Page 2
Problems 3-12

In this section we consider \(8\) classes of integrals with trigonometric functions. Special transformations and subtitutions 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 on problem description to see 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

\[
{\int {{\cos^5}xdx} }
= {\int {{{\left( {{\cos^2}x} \right)}^2}\cos xdx} }
= {\int {{{\left( {1 – {{\sin }^2}x} \right)}^2}\cos x dx} }
= {\int {{{\left( {1 – {u^2}} \right)}^2}du} }
= {\int {\left( {1 – 2{u^2} + {u^4}} \right)du} }
= {{u – \frac{{2{u^3}}}{3} + \frac{{{u^5}}}{5} }+{ C }}
= {{\sin x – \frac{{2{{\sin }^3}x}}{3} }+{ \frac{{{{\sin }^5}x}}{5} }+{ C.}}
\]
Page 1
Problems 1-2
Page 2
Problems 3-12