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

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int e^x \cos x \, dx$. Using integration by parts,$\int u \, dv = uv - \int v \, du$. Let $u = \cos x$ and $dv = e^x \, dx$. Then $du = -\sin x \, dx$ and $v = e^x$. $I = e^x \cos x - \int e^x (-\sin x) \, dx = e^x \cos x + \int e^x \sin x \, dx$. Now,evaluate $\int e^x \sin x \, dx$ using integration by parts again. Let $u = \sin x$ and $dv = e^x \, dx$. Then $du = \cos x \, dx$ and $v = e^x$. $\int e^x \sin x \, dx = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - I$. Substituting this back into the equation for $I$: $I = e^x \cos x + e^x \sin x - I$. $2I = e^x (\cos x + \sin x)$. $I = \frac{e^x (\cos x + \sin x)}{2} + c$.

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

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