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6) Find the integrals of the functions \sin x \sin 2x \sin 3x

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best_answer

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

we can write the integral as follows

\int \sin x.\sin 2x\sin 3x\ dx = \int \sin x\frac{1}{2}[\cos x-\cos 5x]dx
                                                   \\=\frac{1}{2} \int [\sin x.\cos x-\sin x.\cos 5x]dx\\ =\frac{1}{2}\int \frac{\sin 2x}{2}dx-\frac{1}{2}\int \sin x. \cos 5x\ dx\\ =-\frac{\cos 2x}{8}-\frac{1}{4}\int[\sin 6x -\sin 4x]\\ =-\frac{\cos 2x}{8}-\frac{1}{4}[\frac{-\cos 6x}{6}+\frac{\cos 4x}{4}]\\ =-\frac{\cos 2x}{8}+\frac{\cos 6x}{24}-\frac{\cos 4x}{16}+C

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manish

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