Mathematics

# Solve $\int\sin 4x\sin 8xdx$

##### SOLUTION

Consider the given integral.

$I=\int{\sin 4x\sin 8xdx}$

$I=\int{\sin 8x\sin 4xdx}$

We know that

$2\sin A\sin B=\cos \left( A-B \right)-\cos \left( A+B \right)$

Therefore,

$I=\dfrac{1}{2}\int{2\sin 8x\sin 4xdx}$

$I=\dfrac{1}{2}\int{\left( \cos \left( 8x-4x \right)-\cos \left( 8x+4x \right) \right)dx}$

$I=\dfrac{1}{2}\int{\left( \cos 4x-\cos 12x \right)dx}$

$I=\dfrac{1}{2}\left[ \dfrac{\sin 4x}{4}-\dfrac{\sin 12x}{12} \right]+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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