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(B) Let $I = \int e^x \sin x \, dx$.Using integration by parts,$\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.Taking $u = \sin x$ an

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$\frac{1}{2} e^x (\sin x - \cos x) + c$

Vedclass · Answer

(B) Let $I = \int e^x \sin x \, dx$.Using integration by parts,$\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$.Taking $u = \sin x$ and $v = e^x$,we get:$I = \sin x \cdot e^x - \int \cos x \cdot e^x \, dx$.Applying integration by parts again to $\int e^x \cos x \, dx$ with $u = \cos x$ and $v = e^x$:$\int e^x \cos x \, dx = \cos x \cdot e^x - \int (-\sin x) \cdot e^x \, dx = e^x \cos x + \int e^x \sin x \, dx$.Substituting this back into the equation for $I$:$I = e^x \sin x - (e^x \cos x + I) + c$.$I = e^x \sin x - e^x \cos x - I + c$.$2I = e^x (\sin x - \cos x) + c$.$I = \frac{1}{2} e^x (\sin x - \cos x) + c$.

$\int e^x \sin x \, dx = $

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