Find the integrals of the function sin 4x sin 8x - Vidyadip

Question

Find the integrals of the function sin 4x sin 8x

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Answer

$\int \sin 4 x \sin 8 x d x$ = $\int \frac{1}{2}\left\{\ cos (4 x-8 x)-\cos (4 x+8 x)\right\} d x$ , {As Sin(A)Sin(B)=Cos(A-B)-Cos(A+B)}
$= \frac{1}{2} \int(\cos (-4 x)-\cos 12 x d x$
$= \frac{1}{2} \int\{\cos 4 x-\cos 12 x\} d x$
$= \frac{1}{2}\left[\frac{\sin 4 x}{4}-\frac{\sin 12 x}{12}\right]+C$

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