Browse Questions

Find the integrals of the functions$\sin^4x$

Toolbox:
• $(i)1-\sin^2x=\cos^2x.$
• $(ii)\int\cos xdx=\sin x+c.$
• $(iii)\sin^2x=\frac{(1-\cos 2x)}{2}.$
• $(iv)\cos^2x=\frac{(1+\cos 2x)}{2}.$
Given:$I=\int\sin^4xdx.$

This can be written as,

$I=\int(\sin^2x)^2dx$.

But $\sin^2x=\frac{(1-\cos 2x)}{2}.$

$\;\;\;=\int\bigg(\frac{1-\cos 2x}{2}\bigg)\bigg(\frac{1-\cos 2x}{2}\bigg)dx.$

$\;\;\;=\frac{1}{4}\int(1-\cos2x)^2dx.$

This is the form $(a-b)^2=a^2-2ab+b^2.$

$\;\;\;=\frac{1}{4}\int(1+\cos^22x-2\cos2x)dx.$

But $\cos^22x=\frac{(1+\cos 4x)}{2}$.

$\;\;\;=\frac{1}{4}\int[1+\bigg(\frac{1+\cos 4x}{2}\bigg)-2\cos 2x]dx.$

On separating the terms we get,

$1+\frac{1}{2}=\frac{3}{2}.$

$I=\frac{1}{4}\int[\frac{3}{2}+\frac{1}{2}\cos 4x-2\cos 2x]dx.$

on separating terms we get,

$\;\;\;=\frac{3}{8}\int dx+\frac{1}{8}\int \cos 4xdx-\frac{1}{2}\cos 2xdx.$

On integrating we get,

$\;\;\;=\frac{3}{8}x+\frac{1}{8}\frac{1}{4}\sin 4x-\frac{1}{2}\frac{1}{2}\sin 4x+c.$

On multiply

$\;\;\;=\frac{3}{8}x+\frac{1}{32}\sin 4x-\frac{1}{4}\sin 4x+c.$

Taking $\frac{1}{8}$ as a common factor

$\;\;\;=\frac{1}{8}[3x+\frac{1}{4}\sin 4x-2\sin 4x]+c.$