Question

$\int \frac{1}{\sin (x-a)\sin (x-b)}dx=$

• A. $\frac{1}{\sin (b-a)\log \left|{\displaystyle \frac{\sin (x+b)}{\sin (x+a)}}\right|}+C$
• B. $\frac{1}{\sin (b+a)\log \left|{\displaystyle \frac{\sin (x-b)}{\sin (x-a)}}\right|}-C$
• C. $\frac{1}{\sin (b-a)\log \left|{\displaystyle \frac{\sin (x-a)}{\sin (x-b)}}\right|}+C$
• D. None of these

Answer 1

The correct option is C

$\frac{1}{\sin (b-a)\log \left|{\displaystyle \frac{\sin (x-a)}{\sin (x-b)}}\right|}+C$

Explanation for the correct answer:

Finding the value of the given integral:

Given that,

$\int \frac{1}{\sin (x-a)\sin (x-b)}dx$

Multiply and divide by $\sin (a-b)$

$=\frac{1}{\sin (a-b)}\int \frac{\sin (a-b)}{\sin (x-a)\sin (x-b)}dx$

Adding and subtracting $x$ in the numerator

$=\frac{1}{\sin (a-b)}\int \frac{\sin (x-x+a-b)}{\sin (x-a)\sin (x-b)}dx$

$$\begin{gathered} =\frac{1}{\sin (a-b)}\int \frac{\sin \left(\right(x-b)-(x-a\left)\right)}{\sin (x-a)\sin (x-b)}dx \\ =\frac{1}{\sin (a-b)}\int \frac{\sin (x-b)\cos (x-a)-\cos (x-b)\sin (x-a)}{\sin (x-a)\sin (x-b)}dx \end{gathered}$$

Since $Sin(A-B)=\sin A\cos B-CosA\sin B$

$$\begin{gathered} =\frac{1}{\sin (a-b)}\int \frac{\cos (x-a)}{\sin (x-a)}-\frac{\cos (x-b)}{\sin (x-b)}dx \\ =\frac{1}{\sin (a-b)}\left[\int \frac{\cos (x-a)}{\sin (x-a)}dx-\int \frac{\cos (x-b)}{\sin (x-b)}dx\right] \end{gathered}$$

Substituting,

$$\begin{gathered} t_1=\sin (x-a) \\ d{t}_1=\cos (x-a)dx \\ {t}_2=\sin (x-b) \\ d{t}_2=\cos (x-b)dx \end{gathered}$$

$$\begin{gathered} =\frac{1}{\sin (a-b)}\left[\int \frac{d{t}_1}{t_1}-\int \frac{d{t}_2}{t_2}dx\right]\left[\because \int \frac{1}{x}dx=\log \left|x\right|+c\right] \\ =\frac{1}{\sin (a-b)}\left[\log \left|t_1\right|-\log \left|t_2\right|\right]+C \\ =\frac{1}{\sin (a-b)}\log \left|\frac{t_1}{t_2}\right|+C\left[\because \log a-\log b=\log \frac{a}{b}\right] \\ =\frac{1}{\sin (b-a)}\log \left|\frac{\sin (x-a)}{\sin (x-b)}\right|+C \end{gathered}$$

Hence, option (C) is the correct answer.