Question

$\int \frac{\sin x+\cos x}{e^{-x}+\sin x}dx$ is equal to

• A. $\log |1-e^x\sin x|+C$
• B. $\log |1+e^{-x}\sin x|+C$
• C. $\log |1+e^x\sin x|+C$
• D. $\log |1-e^{-x}\sin x|+C$
• E. $\log |1+e^{2x}\sin x|+C$

Answer 1

The correct option is C $\log |1+e^x\sin x|+C$

Let $l=\int \frac{\sin x+\cos x}{e^{-x}+\sin x}dx$
$=\int \frac{e^x(\sin x+\cos x)}{(1+e^x\sin x)}dx$
Put $t=1+e^x\sin x$
$\Rightarrow dt=(e^x\sin x+e^x\cos x)dx$
$=e^x(\sin x+\cos x)dx$
Then, $l=\int \frac{dt}{t}=\log |t|+C$
$\Rightarrow l=\log |1+e^x\sin x|+C$