Question

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

• A. $\frac{-{\cot }^{n}x}{n}+c$
• B. $\frac{-{\cot }^{n}x}{n+1}+c$
• C. $\frac{{\cot }^{n}x}{n}+c$
• D. $\frac{{\cot }^{n}x}{n+1}+c$

Answer 1

The correct option is A $\frac{-{\cot }^{n}x}{n}+c$

Let $I=\int \frac{{\left(\cos x\right)}^{n-1}}{{\left(\sin x\right)}^{n+1}}dx$

$=\int \frac{{\left(\cos x\right)}^{n-1}}{{\left(\sin x\right)}^{n-1}}\times \frac{1}{{\sin }^{2}x}.dx$

$=\int {\left(\cot x\right)}^{n-1}.{\text{cosec}}^{2}x.dx.$

We know that, $d\cot x=-{\text{cosec}}^{2}x.dx$

$\therefore I=-\int {\left(\cot x\right)}^{n-1}.d\left(\cot x\right)$

$=-\frac{{\left(\cot x\right)}^n}{n}+c$

$=-\frac{{\cot }^{n}x}{n}+c$