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Q2  Find the integrals of the functions \sin 3x \cos 4x

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Using identity \sin A\cos B = 1/2[sin(A+B)+sin(A-B)]

, therefore the given integral can be written as

\int \sin 3x\cos 4x=\frac{1}{2}\int sin(3x+4x)+sin(3x-4x)\ dx

                                  =\frac{1}{2}\int sin(7x)-sin(x)\ dx\\ =1/2[\int \sin (7x) dx-\int \sin x\ dx]\\ =\frac{1}{2}[(-1/7)\cos 7x+\cos x+ C]\\ = \frac{\cos x}{2}-\frac{\cos 7x}{14}+C

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manish

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