Question

$\int e^{\sin x}\cdot (\frac{\sin x+1}{\sec x})dx$ is equal to

• A. $\sin x\cdot e^{\sin x}+C$
• B. $\cos x\cdot e^{\sin x}+C$
• C. $e^{\sin x}+C$
• D. $e^{\sin x}(\sin x+1)+C$

Answer 1

The correct option is A $\sin x\cdot e^{\sin x}+C$

$I=\int e^{\sin x}\cdot (\frac{\sin x+1}{\sec x})dx$
$=\int e^{\sin x}(\sin x\cos x+\cos x)dx$
$=\int e^{\sin x}(\sin x+1)\cos xdx$
$\text{Put}\sin x=t\Rightarrow \cos xdx=dt$
$\text{Then},I=\int e^t(t+1)dt$
$=(t+1)e^t-\int e^{t}dt$
$=(t+1)e^t-e^t+C=t{e}^t+C=\sin x\cdot e^{\sin x}+C$