int sin 2x cos 3x , dx = | vedclass

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(B) To solve the integral $\int \sin 2x \cos 3x \, dx$,we use the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$.Here,$A = 2x$ and $

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$\frac{1}{2} \left( \cos x - \frac{1}{5} \cos 5x \right) + c$

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(B) To solve the integral $\int \sin 2x \cos 3x \, dx$,we use the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$.Here,$A = 2x$ and $B = 3x$,so $2 \sin 2x \cos 3x = \sin(2x+3x) + \sin(2x-3x) = \sin 5x + \sin(-x) = \sin 5x - \sin x$.Substituting this into the integral:$\int \sin 2x \cos 3x \, dx = \frac{1}{2} \int 2 \sin 2x \cos 3x \, dx$$= \frac{1}{2} \int (\sin 5x - \sin x) \, dx$$= \frac{1}{2} \left( \int \sin 5x \, dx - \int \sin x \, dx \right)$$= \frac{1}{2} \left( -\frac{\cos 5x}{5} - (-\cos x) \right) + c$$= \frac{1}{2} \left( \cos x - \frac{1}{5} \cos 5x \right) + c$.

$\int \sin 2x \cos 3x \, dx = $

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