KCET · Maths · Indefinite Integration
\(\int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} d x(\) where, \(n \neq 0)\) is equal to
- A \(\frac{\cot ^{n} x}{n}+C\)
- B \(\frac{-\cot ^{n-1} x}{n-1}+C\)
- C \(\frac{-\cot ^{n} x}{n}+C\)
- D \(\frac{\cot ^{n-1} x}{n-1}+C\)
Answer & Solution
Correct Answer
(C) \(\frac{-\cot ^{n} x}{n}+C\)
Step-by-step Solution
Detailed explanation
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 \(\quad t=\cot x\)
\(\Rightarrow \quad d t=-\operatorname{cosec}^{2} x d x=\int t^{n-1}(-d t)\)
\(=-\left(\frac{t^{n}}{n}\right)+C\)
\(=-\frac{1}{n} \cot ^{n} x+C \quad(\because t=\cot x)\)
\(\begin{aligned} I &=\int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} d x, \quad(n \neq 0) \\ &=\int \frac{\cos ^{n-1} x}{\sin ^{n+1} x} \times \frac{\sin ^{2} x}{\sin ^{2} x} d x \\ &=\int \frac{\cos ^{n-1} x}{\sin ^{n-1} x} \cdot \operatorname{cosec}^{2} x d x \\ &=\int \cot ^{n-1} x \cdot \operatorname{cosec}^{2} x d x \end{aligned}\)
Let \(\quad t=\cot x\)
\(\Rightarrow \quad d t=-\operatorname{cosec}^{2} x d x=\int t^{n-1}(-d t)\)
\(=-\left(\frac{t^{n}}{n}\right)+C\)
\(=-\frac{1}{n} \cot ^{n} x+C \quad(\because t=\cot x)\)
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