Question

$\int \frac {\cos^{n-1}x}{\sin^{n+1}x}dx, (where n \, \neq \, 0$ is equal to

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

Answer 1

Answer: (C)

Let $ I =\int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} d x_{1} \,\,\,\,\,\,(n \neq 0) $
$=\int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} \times \frac{\sin ^{2} x}{\sin ^{2} x} d x $
$ =\int \frac{\cos ^{n-1} x}{\sin ^{n-1} x} \cdot \,cosec\,{}^{2} x \,d x $
$ =\int \cot ^{n-1} x \cdot \,cosec\,{}^{2} x \,d x $
Let $ t =\cot x $
$\Rightarrow d t =- cosec\,{}^{2} x\, d x=\int t^{n-1}(-d t) $
$ =-\left(\frac{t^{n}}{n}\right)+C $
$ =-\frac{1}{n} \cot ^{n} x+C \,\,\,\,\,(\because t=\cot x)$