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(B) To evaluate the integral $\int \sin 2x \cos 3x \, dx$,we use the trigonometric identity: $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$. He

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$-\frac{1}{10} \cos 5x + \frac{1}{2} \cos x + C$

Vedclass · Answer

(B) To evaluate the integral $\int \sin 2x \cos 3x \, dx$,we use the trigonometric identity: $\sin A \cos B = \frac{1}{2} [\sin(A+B) + \sin(A-B)]$. Here,$A = 2x$ and $B = 3x$. Substituting these into the identity,we get: $\sin 2x \cos 3x = \frac{1}{2} [\sin(2x+3x) + \sin(2x-3x)] = \frac{1}{2} [\sin 5x + \sin(-x)]$. Since $\sin(-x) = -\sin x$,the expression becomes: $\sin 2x \cos 3x = \frac{1}{2} [\sin 5x - \sin x]$. Now,integrate both sides: $\int \sin 2x \cos 3x \, dx = \frac{1}{2} \int (\sin 5x - \sin x) \, dx$. $= \frac{1}{2} [\int \sin 5x \, dx - \int \sin x \, dx]$. $= \frac{1}{2} [-\frac{1}{5} \cos 5x - (-\cos x)] + C$. $= -\frac{1}{10} \cos 5x + \frac{1}{2} \cos x + C$.

Find: $\int \sin 2x \cos 3x \, dx$

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