Question

${\int }^{}\frac{1+{\cos }^{2}x}{{\sin }^{2}x}dx$

Answer 1

Consider the given integral.

$I=\int \frac{1+{\cos }^{2}x}{{\sin }^{2}x}dx$

$I=\int (\frac{1}{{\sin }^{2}x}+\frac{{\cos }^{2}x}{{\sin }^{2}x})dx$

$I=\int ({\csc }^{2}x+{\cot }^{2}x)dx$

$I=\int ({\csc }^{2}x+{\csc }^{2}x-1)dx$

$I=\int (2{\csc }^{2}x-1)dx$

$I=2(-\cot x)-x+C$

$I=-2\cot x-x+C$

Hence, this is the answer.