int_0^{pi /2} e^x sin x , dx = | vedclass

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Question

(B) Let $I = \int_0^{\pi /2} e^x \sin x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \sin x$ and $dv = e^x \, dx$.

Vedclass · Accepted Answer

$\frac{1}{2}(e^{\pi /2} + 1)$

Vedclass · Answer

(B) Let $I = \int_0^{\pi /2} e^x \sin x \, dx$. 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 $v = e^x$. $I = [e^x \sin x]_0^{\pi /2} - \int_0^{\pi /2} e^x \cos x \, dx$. Applying integration by parts again to $\int e^x \cos x \, dx$: Let $u = \cos x$ and $dv = e^x \, dx$. Then $du = -\sin x \, dx$ and $v = e^x$. $I = [e^x \sin x]_0^{\pi /2} - ([e^x \cos x]_0^{\pi /2} - \int_0^{\pi /2} e^x (-\sin x) \, dx)$. $I = [e^x \sin x]_0^{\pi /2} - [e^x \cos x]_0^{\pi /2} - I$. $2I = [e^x(\sin x - \cos x)]_0^{\pi /2}$. $2I = (e^{\pi /2}(\sin(\pi /2) - \cos(\pi /2))) - (e^0(\sin 0 - \cos 0))$. $2I = (e^{\pi /2}(1 - 0)) - (1(0 - 1))$. $2I = e^{\pi /2} + 1$. $I = \frac{1}{2}(e^{\pi /2} + 1)$.

$\int_0^{\pi /2} e^x \sin x \, dx = $

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