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

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

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(A) We have,$I = \int e^{\sin x} \sin 2x \, dx$ Using the identity $\sin 2x = 2 \sin x \cos x$,we get: $I = \int e^{\sin x} (2 \sin x \cos x) \, dx = 2 \int e^{\sin x} \sin x \cos x \, dx$ Let $\sin x = t$,then $\cos x \, dx = dt$. Substituting these into the integral: $I = 2 \int e^t \cdot t \, dt$ 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: $I = 2 e^{\sin x} (\sin x - 1) + C$

The value of $\int e^{\sin x} \sin 2 x \, dx$ is

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