# Integrate : $\int e^x [\cot x + \log \sin x]dx$

$(a)\;e^x \log \sin x+c \qquad(b)\;e^x \log \cos x+c \qquad(c)\;e^x \log \tan x+c \qquad (d)\;None$

$\int e^x. \cot x dx+\int e^x \log \sin x dx$
$I= \int e^x \cot x dx+ \int e^x \log \sin x dx$
$I= \int e^x \cot x dx+ \int e^x \log \sin x dx- \int\cot x.e^{x} dx$
=>$e^x \log \sin x+c$
Hence a is the correct answer.