Question

The value of $\int e^{sinx}sin2xdx$ is

• A. $2e^{sinx}(sinx-1)+C$
• B. $2e^{sinx}(sinx+1)+C$
• C. $2e^{sinx}(cosx+1)+C$
• D. $2e^{sinx}(cosx-1)+C$

Answer 1

Answer: (A)

$\left(I=\int e^{sinx}sin2xdx=2\int e^{sinx}sinxcosxdx\right)$
Let $t=\sin x$
$\Rightarrow dt=\cos xdx$
Therefore, $I=2\int {te}^{t}dt$
$=2\left(t{e}^t-e^t\right)+C$
$=2(\sin x-1)e^{\sin x}+C$