Browse Questions

# Find the integrals of the functions$\sin4x\;\sin8x$

Toolbox:
• (i)$2\sin x\sin y=\cos(A-B)-\cos(A+B).$
• $(ii)\int\sin xdx=-\cos x+c.$
• $(iii)\int\cos xdx=\sin x+c.$
Given $I=\int \sin 4x\sin 8x.$

Using the information from the tool box we write as,

$I=\frac{1}{2}\int[\cos(4x-8x)-\cos(4x+8x)]dx.$

$\;\;\;=\frac{1}{2}\int (\cos 4x-\cos 12x)dx.$

On separating the terms

$\;\;\;=\frac{1}{2}\int\cos 4xdx-\frac{1}{2}\int\cos 12x dx.$

On integrating we get,

$\;\;\;=\frac{1}{2}[\frac{1}{4}\sin 4x]-\frac{1}{2}[\frac{1}{12}\sin 12x]+c.$

Taking $\frac{1}{4}$ as the common factor we get,

$\;\;\;=\frac{1}{4}\left\{[\frac{1}{2}\sin4x]-[\frac{1}{3}\sin 12x]\right\}+c$