int frac{e^{sin x}(sin 2x - 8 cos x)}{2(sin x | vedclass

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Question

(D) Let $I = \int \frac{e^{\sin x}(\sin 2x - 8 \cos x)}{2(\sin x - 3)^2} dx$. Using $\sin 2x = 2 \sin x \cos x$,we have: $I = \int \frac{e^{\sin x}(2

Vedclass · Accepted Answer

$\frac{e^{\sin x}}{\sin x - 3} + c$

Vedclass · Answer

(D) Let $I = \int \frac{e^{\sin x}(\sin 2x - 8 \cos x)}{2(\sin x - 3)^2} dx$. Using $\sin 2x = 2 \sin x \cos x$,we have: $I = \int \frac{e^{\sin x}(2 \sin x \cos x - 8 \cos x)}{2(\sin x - 3)^2} dx = \int \frac{e^{\sin x} \cos x (\sin x - 4)}{(\sin x - 3)^2} dx$. Let $t = \sin x$,then $dt = \cos x dx$. $I = \int \frac{e^t (t - 4)}{(t - 3)^2} dt = \int e^t \left( \frac{t - 3 - 1}{(t - 3)^2} \right) dt = \int e^t \left( \frac{1}{t - 3} - \frac{1}{(t - 3)^2} \right) dt$. Using the formula $\int e^t (f(t) + f'(t)) dt = e^t f(t) + c$,where $f(t) = \frac{1}{t - 3}$ and $f'(t) = -\frac{1}{(t - 3)^2}$. Thus,$I = e^t \left( \frac{1}{t - 3} \right) + c = \frac{e^{\sin x}}{\sin x - 3} + c$.

$\int \frac{e^{\sin x}(\sin 2x - 8 \cos x)}{2(\sin x - 3)^2} dx =$

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