Question

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

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

Answer 1

Answer: (A)

Let I = $\int e^{\sin x} \left[\frac{\sin x+1}{\sec x}\right]dx = \int e^{\sin x} \left[\sin x+1\right] \cos\, x\, dx$
Put $\sin x = t$
$ \Rightarrow \cos \,x \,dx =dt $
$ \therefore \,I = \int e^{t} \left[1+t\right]dt =e^{t} t +c =e^{\sin x} . \sin x +c $