Question

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

• A. $\log \left|\cos \left(x{e}^x\right)\right|+C$
• B. $\log \left|\cot \left(x{e}^x\right)\right|+C$
• C. $\log \left|\sec \left(x{e}^{-x}\right)\right|+C$
• D. $\log \left|\cos \left(x{e}^{-x}\right)\right|+C$
• E. $\log \left|\sec \left(x{e}^x\right)\right|+C$

Answer 1

Answer: (C)

$\log \left|\sec \left(x{e}^{-x}\right)\right|+C$

Let $I=\int \frac{(1+x)e^x}{\cot \left(x{e}^x\right)}dx$
Put $x{e}^x=t$
$\Rightarrow (x{e}^x+e^x)dx=dt$
$\Rightarrow (x+1)e^{x}dx=dt$
$\therefore I=\int \frac{dt}{\cot t}$
$=\int \tan tdt$
$=\log \left|\sec t\right|+C$
Again, put $t=x{e}^x$
$\Rightarrow I=\log \left|\sec \left(x{e}^x\right)\right|+C$