Q64 / 180 · KCET 2013 · Mathematics

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

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

Solution

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

Hence, the correct answer is option: 'C'.

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