Question

$\int e^{2x}\left(\frac{\cot 2x-1}{\cos x\sin x}\right)dx=$

• A. 2 $e^{2x}\csc 2x+c$
• B. $e^{2x}\cot 2x+c$
• C. $-e^{2x}cosec2x+c$
• D. $e^{2x}cosec2x+c$

Answer 1

The correct option is C $-e^{2x}cosec2x+c$

$\int e^{2x}\left(\frac{\cot 2x-1}{2\cos x\sin x}\right)d2x$
$\int e^{2x}(\cot 2xcosec2x-cosec2x)d2x$
$2x=t$
$\int e^t(\cot tcosect-cosect)dt$
$=-\int e^t(cosect-\cot tcosect)dt$
It is in the form of $\int e^x\left[f\right(x)+f^1(x\left)\right]dx=e^{x}f\left(x\right)+c$
$=-e^{t}cosect+c$
$=-e^{2x}cosec2x+c$