Question

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

• A. $-\cot x-2x+c$
• B. $-2\cot x-2x+c$
• C. $-2\cot x-x+c$
• D. $-2\cot x+x+c$

Answer 1

The correct option is C $-2\cot x-x+c$

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

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

$=\int \left(\cos e{c}^{2}x+{\cot }^{2}x\right)dx$

$=\int \left(\cos e{c}^{2}x+\cos e{c}^{2}x-1\right)dx$

$=2\int \cos e{c}^{2}xdx-\int dx$

$=-2\cot x-x+c$