physicscatalyst.com logo




Integration Using Trigonometric Identities




Integration Using Trigonometric Identities

We can do Integration Using Trigonometric Identities for some functions.

A. $\int \cos (Ax) \cos (Bx) \; dx$
We know that
$\cos(A+B)=\cos(A)\cos(B)- \sin(A)\sin(B)$
$\cos(A-B)=\cos(A)\cos(B)+ \sin(A)\sin(B)$
Therefore
$\cos(A)\cos(B) = \frac {1}{2} [\cos(A+B) + \cos(A-B)]$
So we can convert into sum of cos function which is easy to integrate

Example
$\int \cos (x) \cos (2x) \; dx$
$=\frac {1}{2} \int [\cos 3x + \cos x] dx= \frac {1}{2} [ - \frac {1}{3} \sin(3x) - \sin(x)] + C$

B. $\int \sin (Ax) \sin (Bx) \; dx$
We know that
$\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$
$\cos(A-B)=\cos(A)\cos(B)+\sin(A)\sin(B)$
Therefore
$\sin(A)\sin(B) = \frac {1}{2} [\cos(A+B) - \cos(A-B)]$
So we can convert into difference of cos function which is easy to integrate

Example
$\int \sin (2x) \sin (x) \; dx$
$=\frac {1}{2} \int [\cos(3x) - \cos(x)] dx= \frac {1}{2} [ - \frac {1}{3} \sin(3x) + \sin(x)] + C$

C. $\int \sin (Ax) \cos (Bx) \; dx$
We know that
$\sin(A+B)=\sin(A)\cos(B)+ \sin(B)\cos(A)$
$\sin(A-B)=\sin(A)\cos(B)- \sin(B)\cos(A)$
Therefore
$\sin(A) \cos(B) = \frac {1}{2} [\sin(A+B) + \sin(A-B)]$
So we can convert into sum of sin function which is easy to integrate

Example
$\int \sin (2x) \cos (3x) \; dx$
$=\frac {1}{2} \int [\sin 5x - \sin x] dx= \frac {1}{2} [ - \frac {1}{5} \cos 5x - \cos x] + C$

D. $\int \sin^2 x \; dx$ or $\int \cos^2 x \; dx$
We know that
$\cos2x=\cos^{2}x-\sin^{2}x=2\cos^{2}x-1=1-2\sin^{2}x}$
So
$\cos^2 x= \frac {1+ \cos 2x}{2}$
$\sin^2 x= \frac {1- \cos 2x}{2}$

E.$\int \sin^3 x \; dx$ or $\int \cos^3 x \; dx$
We know that
$\sin(3x)=3\sin(x)-4\sin^{3}x$
$\sin^{3}x= \frac { 3 \sinx - \sin 3x}{4}$
Also
$\cos(3x)=4\cos^{3}x-3\cos(x)$
$\cos^{3}x= \frac {\cos 3x + 3 \cos x}{4}$

F. $\int \tan^2 x \; dx$ or $\int \cot^2 x \; dx$
We know that
$\sec^2x =1 + \tan^2x$
or $\tan^2x =\sec^2 x -1$
$\int \tan^2 x dx = \int (\sec^2 x -1) dx$
Now $\int ( \sec^2 x) \; dx = \tan x + C$
Therefore
$\int \tan^2 x \; dx = \int (\sec^2 x -1) dx= \tan x -x + C$
We know that
$cosec^2 x =1 + \cot^2 x$

or $ \cot^2 x = cosec^2 x -1$
$\int \cot^2 x \; dx = \int (cosec^2 x -1) \; dx$
Now $\int ( cosec^2 x) \; dx = -\cot x + C$
Therefore
$\int \cot^2 x \; dx = \int (cosec^2 x -1) \; dx= -\cot x -x + C$

Solved examples

Example 1
$\int \sin^4 x \; dx$
Solution
$\int \sin^4 x \; dx = \int (\sin^2 x)^2 \; dx=\int (\frac {1-\cos 2x}{2})^2 \; dx $
$=\frac {1}{4} \int [ 1 + \cos^2 2x - 2\cos 2x] \; dx$
$=\frac {1}{4} \int [ 1 + \frac {1+ \cos 4x}{2} - 2\cos 2x] \; dx$
$= \frac {1}{4} \int [ 3/2 + \cos 4x/2 -2 \cos 2x] \; dx$
$= \frac {3}{8} x + \frac {1}{32} \sin 4x -\frac {1}{4} \sin 2x + C$


Also Read




Go back to Class 12 Main Page using below links
Class 12 Maths Class 12 Physics Class 12 Chemistry Class 12 Biology


Latest Updates
Synthetic Fibres and Plastics Class 8 Practice questions

Class 8 science chapter 5 extra questions and Answers

Mass Calculator

3 Fraction calculator

Garbage in Garbage out Extra Questions7