ExamBro
ExamBro
MHT CET · Maths · Indefinite Integration

If \(\mathrm{I}=\int \frac{\mathrm{d} x}{\sin (x-\mathrm{a}) \sin (x-\mathrm{b})}\), then \(\mathrm{I}\) is given by

  1. A \(\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \log |\sin (x-\mathrm{a}) \sin (x-\mathrm{b})|+\mathrm{c}\) where \(\mathrm{c}\) is a constant of integration.
  2. B \(\log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+c\), where \(\mathrm{c}\) is a constant of integration.
  3. C \(\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \log \left|\frac{\sin (x-\mathrm{a})}{\sin (x-\mathrm{b})}\right|+\mathrm{c}\), where \(\mathrm{c}\) is a constant of integration.
  4. D \(\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \log \left|\frac{\sin (x-\mathrm{b})}{\sin (x-\mathrm{a})}\right|+\mathrm{c}\), where \(\mathrm{c}\) is \(\mathrm{a}\) constant of integration.
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{1}{\sin (\mathrm{b}-\mathrm{a})} \log \left|\frac{\sin (x-\mathrm{b})}{\sin (x-\mathrm{a})}\right|+\mathrm{c}\), where \(\mathrm{c}\) is \(\mathrm{a}\) constant of integration.

Step-by-step Solution

Detailed explanation

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