^4x - NCERT Class XII Mathematics Chapter - - Solutions

NCERT Class XII Mathematics Chapter - - Solutions

Exercise - 7.3
Find the integrals of the functions in Exercises 62 to 83.

Question : 71 of 271

Marks: +1, -0

\sin^4x

Solution:

Let

I=\int\sin^4xdx

=\int\left(\frac{1-\cos2x}{2}\right)^2dx

(\because \cos2x = 1 - 2\sin^2 \sin x)

=\frac{1}{4}\int(1+\cos^2 2x-2\cos2x)dx

=\frac{1}{4}\int\left[1+\frac{1+\cos4x}{2}-2\cos 2x\right]dx\;\;(\because \cos2x = 2\cos2x - 1)

=\frac{1}{4}\int dx + \frac{1}{8}\int(1+\cos4x)dx - \frac{2}{4}\int\cos2xdx

=\frac{3}{8}\int dx + \frac{1}{8}\int\cos4x dx - \frac{1}{2}\int\cos2xdx

=\frac{3}{8}x + \frac{1}{32}\sin4x - \frac{1}{4}\sin2x + C

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