Mathematics

$$\displaystyle\int \frac{\cos ^{2}x}{\sin ^{4}x}dx.$$


ANSWER

$$\displaystyle -\frac{1}{3}\cot ^{3}x.$$


SOLUTION
Let $$\displaystyle I=\int  \frac { \cos ^{ 2 } x }{ \sin ^{ 4 } x } dx=-\int { \cot ^{ 2 }{ x } \csc ^{ 2 }{ x }  } dx$$

Put $$\displaystyle \cot { x } =t\Rightarrow -\csc ^{ 2 }{ x } dx=dt$$
Therefore 
$$\displaystyle I=-\int { { t }^{ 2 }dt } =-\frac { { t }^{ 3 } }{ 3 } =-\frac { 1 }{ 3 } \cot ^{ 3 }{ x } $$
Hence, option 'C' is correct.
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