∫e^x sin x dx\no a (e^x cosx - e^x sinx)/2 + c\no b -e^x cosx + e^x sinx + c\no c (e^x cosx + e^x sinx)/2 +…

Explanation:

Step1: Apply integration - by - parts

Let $u = \sin x$, $dv=e^{x}dx$. Then $du=\cos xdx$, $v = e^{x}$. By the integration - by - parts formula $\int u;dv=uv-\int v;du$, we have $\int e^{x}\sin xdx=e^{x}\sin x-\int e^{x}\cos xdx$.

Step2: Apply integration - by - parts again

For $\int e^{x}\cos xdx$, let $u = \cos x$, $dv = e^{x}dx$. Then $du=-\sin xdx$, $v = e^{x}$. So $\int e^{x}\cos xdx=e^{x}\cos x+\int e^{x}\sin xdx$.

Step3: Substitute the result of Step2 into Step1

$\int e^{x}\sin xdx=e^{x}\sin x-(e^{x}\cos x+\int e^{x}\sin xdx)$.

Step4: Solve for $\int e^{x}\sin xdx$

$\int e^{x}\sin xdx=e^{x}\sin x - e^{x}\cos x-\int e^{x}\sin xdx$. Add $\int e^{x}\sin xdx$ to both sides: $2\int e^{x}\sin xdx=e^{x}\sin x - e^{x}\cos x$. Then $\int e^{x}\sin xdx=\frac{e^{x}\sin x - e^{x}\cos x}{2}+C=\frac{-e^{x}\cos x + e^{x}\sin x}{2}+C$.

Answer:

E. $\frac{-e^{x}\cos x + e^{x}\sin x}{2}+C$