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KCET · Maths · Indefinite Integration

\(\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x\) is equal to

  1. A \(\frac{-1}{\sin a} \log |\sin x \operatorname{cosec}(x-a)|+c\)
  2. B \(\frac{-1}{\sin a} \log [\sin (x-a) \sin x]+c\)
  3. C \(\frac{1}{\sin a} \log [\sin (x-a) \operatorname{cosec} x]+c\)
  4. D \(\frac{1}{\sin a} \log [\sin (x-a) \sin x]+c\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{-1}{\sin a} \log |\sin x \operatorname{cosec}(x-a)|+c\)

Step-by-step Solution

Detailed explanation

Let \(I=\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x\)
\(=\int \frac{\sin a}{\sin a \sin (x-a) \sin x} d x\)
\(=-\frac{1}{\sin a} \int \frac{\sin [(x-a)-x]}{\sin (x-a) \sin x} d x\)
\(=-\frac{1}{\sin a} \int\left[\frac{\sin (x-a) \cos x-\cos (x-a) \sin x}{\sin (x-a) \sin x}\right] d x\)
\(=-\frac{1}{\sin a} \int[\cot x-\cot (x-a)] d x\)
\(=-\frac{1}{\sin a}[\log |\sin x|-\log |\sin (x-a)|]+c\)
\(=\frac{-1}{\sin a}[\log |\sin x \operatorname{cosec}(x-a)|]+c\)