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Q3  Find the integrals of the functions \cos 2x \cos 4x \cos 6x

Answers (1)

best_answer

Using identity
 cosAcosB = \frac{1}{2}[cos(A+B)+cos(A-B)]

\int \cos 2x.\cos 4x.\cos 6x = \int \cos 2x. \frac{1}{2}[(\cos 10x)+\cos 2x]dx

Again use the same identity mentioned in the first line

                                                   \\= \frac{1}{2}\int (\cos 2x.\cos 10x+\cos 2x. \cos 2x)dx\\ =\frac{1}{2}\int\frac{1}{2}({\cos12x +\cos 8x})dx+\frac{1}{2}\int (\frac{1+\cos 4x}{2})dx\\ =\frac{\sin 12x}{48}+\frac{\sin 8x}{32}+\frac{\sin 4x}{16}+ x/4+C

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manish

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