KCET · Maths · Indefinite Integration
\(\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x\) is equal to
- A \(\frac{-1}{\sin a} \log |\sin x \operatorname{cosec}(x-a)|+c\)
- B \(\frac{-1}{\sin a} \log [\sin (x-a) \sin x]+c\)
- C \(\frac{1}{\sin a} \log [\sin (x-a) \operatorname{cosec} x]+c\)
- D \(\frac{1}{\sin a} \log [\sin (x-a) \sin x]+c\)
Answer & Solution
Correct Answer
(A) \(\frac{-1}{\sin a} \log |\sin x \operatorname{cosec}(x-a)|+c\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \operatorname{cosec}(x-a) \operatorname{cosec} x d x\)
\(=\int \frac{\sin a}{\sin a \sin (x-a) \sin x} d x\)
\(=-\frac{1}{\sin a} \int \frac{\sin [(x-a)-x]}{\sin (x-a) \sin x} d x\)
\(=-\frac{1}{\sin a} \int\left[\frac{\sin (x-a) \cos x-\cos (x-a) \sin x}{\sin (x-a) \sin x}\right] d x\)
\(=-\frac{1}{\sin a} \int[\cot x-\cot (x-a)] d x\)
\(=-\frac{1}{\sin a}[\log |\sin x|-\log |\sin (x-a)|]+c\)
\(=\frac{-1}{\sin a}[\log |\sin x \operatorname{cosec}(x-a)|]+c\)
\(=\int \frac{\sin a}{\sin a \sin (x-a) \sin x} d x\)
\(=-\frac{1}{\sin a} \int \frac{\sin [(x-a)-x]}{\sin (x-a) \sin x} d x\)
\(=-\frac{1}{\sin a} \int\left[\frac{\sin (x-a) \cos x-\cos (x-a) \sin x}{\sin (x-a) \sin x}\right] d x\)
\(=-\frac{1}{\sin a} \int[\cot x-\cot (x-a)] d x\)
\(=-\frac{1}{\sin a}[\log |\sin x|-\log |\sin (x-a)|]+c\)
\(=\frac{-1}{\sin a}[\log |\sin x \operatorname{cosec}(x-a)|]+c\)
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