int e^{sin x} sin 2x , dx = . . . . . . + | vedclass

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(A) Let $I = \int e^{\sin x} \sin 2x \, dx$. Using the identity $\sin 2x = 2 \sin x \cos x$,we get: $I = \int e^{\sin x} (2 \sin x \cos x) \, dx$. Let

Vedclass · Accepted Answer

$2e^{\sin x}(\sin x - 1)$

Vedclass · Answer

(A) Let $I = \int e^{\sin x} \sin 2x \, dx$. Using the identity $\sin 2x = 2 \sin x \cos x$,we get: $I = \int e^{\sin x} (2 \sin x \cos x) \, dx$. Let $t = \sin x$,then $dt = \cos x \, dx$. Substituting these into the integral: $I = \int e^t (2t) \, dt = 2 \int t e^t \, dt$. Using integration by parts,$\int u \, dv = uv - \int v \, du$,where $u = t$ and $dv = e^t \, dt$: $I = 2 [t e^t - \int e^t \, dt] = 2 [t e^t - e^t] + c$. $I = 2 e^t (t - 1) + c$. Substituting $t = \sin x$ back: $I = 2 e^{\sin x} (\sin x - 1) + c$. Thus,the correct option is $A$.

$\int e^{\sin x} \sin 2x \, dx = $ . . . . . . $+ c$.

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