Question

Let $f\left(x\right)=\frac{2{\sin }^{2}x-1}{\cos x}+\frac{\cos x(2\sin x+1)}{1+\sin x}.$ Then $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx$ equals
(where $C$ is the constant of integration)

• A. $e^x\tan x+C$
• B. $e^x\cot x+C$
• C. $e^x\text{cosec}x+C$
• D. $e^x{\sec }^{2}x+C$

Answer 1

The correct option is A $e^x\tan x+C$

We have,
$f\left(x\right)=\frac{\cos x(1+2\sin x)}{1+\sin x}-\frac{{\cos }^{2}x-{\sin }^{2}x}{\cos x}$
$=\cos x+\frac{\cos x\sin x}{1+\sin x}-\cos x+\frac{{\sin }^{2}x}{\cos x}$
$=\frac{\cos x\sin x}{1+\sin x}+\frac{{\sin }^{2}x}{\cos x}$
$=\frac{\sin x({\cos }^{2}x+{\sin }^{2}x)+{\sin }^{2}x}{\cos x(1+\sin x)}$
$f\left(x\right)=\tan x$

Now, $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx$
$=\int e^{x}f\left(x\right)dx$
$=e^x\tan x+C$