Question

Evaluate $\int e^{\sin x}\sin 2xdx$

• A. $2e^{\sin x}(\sin x+1)+c$
• B. $e^{\sin x}(\sin x+2)+c$
• C. $e^{\sin x}(2\sin x-2)+c$
• D. $e^{\sin x}(3\sin x-2)+c$

Answer 1

The correct option is C $e^{\sin x}(2\sin x-2)+c$

$\int e^{\sin x}\cdot 2\sin x\cos xdx$

$=2\int e^{\sin x}\cdot \sin xd\left(\sin x\right)$

Take $\sin x=t$

$=2\int e^t\cdot tdt$

We know, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

Now by integrating the functions by parts, by taking $t$ as $u$ and $e^t$ as $v$, we get

$=2(t{e}^t-e^t)$

$=2e{}^t(t-1)$

Put the value of $t$

$=2e^{\sin x}(\sin x-1)=e^{\sin x}(2\sin x-2)$

$\Rightarrow \int e{}^{\sin x}\sin 2xdx=e^{\sin x}(2\sin x-2)+c$