Question

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

• A. $cosec(b-a).ln\left|\begin{array}{c}\frac{sin(x-b)}{sin(x-a)}\end{array}\right|+c$
• B. $sec(b-a).ln\left|\begin{array}{c}\frac{sin(x-b)}{sin(x-a)}\end{array}\right|+c$
• C. $sec(b-a).ln\left|\begin{array}{c}\frac{sin(x+b)}{sin(x+a)}\end{array}\right|+c$
• D. $cosec(b-a).ln\left|\begin{array}{c}\frac{sin(x+b)}{sin(x+a)}\end{array}\right|+c$

Answer 1

Answer: (A)

$cosec(b-a).ln\left|\begin{array}{c}\frac{sin(x-b)}{sin(x-a)}\end{array}\right|+c$

Let $I=\int \frac{dx}{\sin (x-a)\sin (x-b)}=\int \csc (a-x)\csc (b-x)dx$
Substitute $u=-x\Rightarrow du=-dx$
$I=-\int \csc (a+u)\csc (b+u)dx=-\int \frac{2}{\cos (a-b)-\cos (a+b+2u)}du$
Substitute $t=a+b+2u\Rightarrow ds=2dt$

$I=-\int \frac{1}{\cos (a-b)-\cot (a+b+2u)}dt$
Substitute $p=\tan \frac{t}{2}\Rightarrow dp=\frac{1}{2}{\sec }^2\frac{t}{2}dt$

$I=-\int \frac{2}{(p^2+1)(\cos (a-b)-\frac{1-p^2}{p^2+1})}dp$
$=-2\int \frac{1}{p^2\cos (a-b)+\cos (a-b)+p^2-1}dp$
$=-\frac{2}{\cos (a-b)-1}\int \frac{1}{\frac{p^2\left(\cos (a-b)+1\right)}{\cos (a-b)-1}+1}dp$
$=-\frac{2\sqrt{-{\tan }^2\left(\frac{a-b}{2}\right)}{\tan }^{-1}\left(\frac{p}{\sqrt{-{\tan }^2\left(\frac{a-b}{2}\right)}}\right)}{\cos (a-b)-1}$
$=\csc (a-b)\left(\log \sin (a-x)-\log \left(\sin (b-x)\right)\right)$

$=\csc (a-b)\log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+c$