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Q7  Find the integrals of the functions \sin 4x \sin 8x

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best_answer

Using identity

sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))

we can write the following integral as 

\sin 4x \sin 8x = 
\\=\frac{1}{2}\int(\cos 4x - \cos 12x) dx\\ =\frac{1}{2} [\int\cos 4x\ dx - \int \cos 12x\ dx]\\ =\frac{\sin 4x}{8}-\frac{\sin 12x}{24}+C

Posted by

manish

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