Question

$\int \text{cosec}(x-a)\text{cosec}xdx=$

• A. $\left(\frac{1}{\sin a}\right)\log \left|\frac{\sin (x-a)}{\sin x}\right|+C$
• B. $\left(\frac{-1}{\sin a}\right)\log \sin (x-a)\sin x+C$
• C. $\left(\frac{1}{\sin a}\right)\log \left(\sin (x-a)+\text{cosec}x\right)+C$
• D. $\left(\frac{1}{\sin a}\right)\log \left(\sin (x-a)\sin x\right)+C$

Answer 1

The correct option is A

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

Explanation for Correct answer:

Finding the value of the given integral:

$$\begin{gathered} \int \text{cosec}(x-a)\text{cosec}xdx=\int \frac{1}{\sin (x-a)\sin x}dx\left[\because \text{cosec}x=\frac{1}{\sin x}\right] \\ =\frac{1}{\sin a}\int \frac{\sin a}{\sin (x-a)\sin x}dx\left[\because \text{multiply and divide by}\sin a\right] \\ =\frac{1}{\sin a}\int \frac{\sin \left(x-x+a\right)}{\sin (x-a)\sin x}dx\left[\because \text{add and subtract}x\text{in numberator}\right] \\ =\frac{1}{\sin a}\int \frac{\sin \left(x-\left(x-a\right)\right)}{\sin (x-a)\sin x}dx \\ =\frac{1}{\sin a}\int \frac{\sin x\cos \left(x-a\right)-\cos x\sin (x-a)}{\sin (x-a)\sin x}dx\left[\because \sin (A-B)=\sin A\cos B-\cos A\sin B\right] \\ =\frac{1}{\sin a}\int \frac{\sin x\cos \left(x-a\right)}{\sin (x-a)\sin x}-\frac{\cos x\sin (x-a)}{\sin (x-a)\sin x}dx \\ =\frac{1}{\sin a}\int \frac{\cos \left(x-a\right)}{\sin (x-a)}-\frac{\cos x}{\sin x}dx \\ =\frac{1}{\sin a}\int \cot \left(x-a\right)-\cot \left(x\right)dx \\ =\frac{1}{\sin a}\left[\log \left|\sin (x-a\right|-\log \left|\sin x\right|\right]+C\left(\because \int \cot x=\log \left|\sin x\right|\right) \\ =\frac{1}{\sin a}\left[\log \left|\frac{\sin (x-a)}{\sin x}\right|\right]+C\left(\because \log a-\log b=\log \frac{a}{b}\right) \end{gathered}$$Hence, option (A) is the correct answer.