Mathematics

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

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

##### 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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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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