Question

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

• A. $\frac{1}{\sin (b-a)}\log |\frac{\sin (x-b)}{\sin (x-a)}|+c$
• B. $\frac{1}{\sin (a-b)}\log |\frac{\sin (x-b)}{\sin (x-a)}|+c$
• C. $\frac{1}{\sin (a+b)}\log |\frac{\sin (x-a)}{\sin (x-b)}|+c$
• D. $\frac{1}{\cos (a+b)}\log |\frac{\sin (x-a)}{\sin (x-b)}|+c$

Answer 1

The correct option is A $\frac{1}{\sin (b-a)}\log |\frac{\sin (x-b)}{\sin (x-a)}|+c$

$\frac{1}{sin(b-a)}\int \frac{sin\left[\right(x-a)-(x-b\left)\right]}{sin(x-a)sin(x-b)}dx$

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

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

$=\frac{1}{sin(b-a)}\left[\log |sin\right(x-b)|-\log |sin(x-a\left)|\right]$

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