If int {e^x sin x , dx} = frac{1}{2} e^x cdot | vedclass

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(A) Let $I = \int e^x \sin x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin x$ and $dv = e^x \, dx$. Then $du =

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$\sin x - \cos x$

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(A) Let $I = \int e^x \sin x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin x$ and $dv = e^x \, dx$. Then $du = \cos x \, dx$ and $v = e^x$.$I = e^x \sin x - \int e^x \cos x \, dx$.Applying integration by parts again to $\int e^x \cos x \, dx$ with $u = \cos x$ and $dv = e^x \, dx$:$I = e^x \sin x - [e^x \cos x - \int e^x (-\sin x) \, dx]$$I = e^x \sin x - e^x \cos x - \int e^x \sin x \, dx$$I = e^x \sin x - e^x \cos x - I$$2I = e^x (\sin x - \cos x) + c$$I = \frac{1}{2} e^x (\sin x - \cos x) + c$.Comparing this with the given expression $\frac{1}{2} e^x \cdot a + c$,we get $a = \sin x - \cos x$.

If $\int {e^x \sin x \, dx} = \frac{1}{2} e^x \cdot a + c$,then $a = $

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