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

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