If $\int \frac{\cos ^4 x}{\sin ^2 x} d x=A \cot x+B \sin 2 x+\frac{C}{2} x+D$, then
Answer & explanation
Correct answer: option 2
We have,
$I =\int \frac{\cos ^4 x}{\sin ^2 x} d x$
$\Rightarrow I =\int \frac{\left(1-\sin ^2 x\right)^2}{\sin ^2 x} d x$
$\Rightarrow I=\int\left(cosec^2 x+\sin ^2 x-2\right) d x$
$\Rightarrow I=-\cot x+\frac{1}{2}\left(x-\frac{\sin 2 x}{2}\right)-2 x+D$
$\Rightarrow I=-\cot x-\frac{1}{4} \sin 2 x-\frac{3}{2} x+D$
Hence, $A=-1, B=-\frac{1}{4}, C=-3$