int frac{sin 2x (1 - frac{3}{2} cos x)}{e^{si | vedclass

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(B) Let $t = \sin^2 x + \cos^3 x$. Then,differentiating with respect to $x$,we get: $dt = (2 \sin x \cos x + 3 \cos^2 x (-\sin x)) \, dx$ $dt = (\sin

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$-e^{-(\sin^2 x + \cos^3 x)} + c$,where $c$ is a constant of integration.

Vedclass · Answer

(B) Let $t = \sin^2 x + \cos^3 x$. Then,differentiating with respect to $x$,we get: $dt = (2 \sin x \cos x + 3 \cos^2 x (-\sin x)) \, dx$ $dt = (\sin 2x - 3 \cos^2 x \sin x) \, dx$ $dt = \sin 2x (1 - \frac{3}{2} \cos x) \, dx$ Substituting this into the integral: $\int \frac{1}{e^t} \, dt = \int e^{-t} \, dt$ $= -e^{-t} + c$ $= -e^{-(\sin^2 x + \cos^3 x)} + c$.

$\int \frac{\sin 2x (1 - \frac{3}{2} \cos x)}{e^{\sin^2 x + \cos^3 x}} \, dx =$

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