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$\frac{d x}{\sin (x-a) \sin (x-b)}$ is equal to
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Verified Answer
The correct answer is:
$\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$
$\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$
Suppose,
$$
\begin{aligned}
&I=\int \frac{d x}{\sin (x-a) \sin (x-b)} \\
&=\frac{1}{\sin (b-a)} \int \frac{\sin (b-a)}{\sin (x-a) \sin (x-b)} d x \\
&=\frac{1}{\sin (b-a)} \int \frac{\sin [(x-a)-(x-b)]}{\sin (x-a) \sin (x-b)} d x \\
&=\frac{1}{\sin (b-a)} \\
&=\frac{1}{\sin (b-a)} \int(\cot (x-b)-\cot (x-a)) d x \\
&=\frac{1}{\sin (b-a)}(\log |\sin (x-b)|-\log |\sin n(x-a)|)+C
\end{aligned}
$$
$=\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$
$$
\begin{aligned}
&I=\int \frac{d x}{\sin (x-a) \sin (x-b)} \\
&=\frac{1}{\sin (b-a)} \int \frac{\sin (b-a)}{\sin (x-a) \sin (x-b)} d x \\
&=\frac{1}{\sin (b-a)} \int \frac{\sin [(x-a)-(x-b)]}{\sin (x-a) \sin (x-b)} d x \\
&=\frac{1}{\sin (b-a)} \\
&=\frac{1}{\sin (b-a)} \int(\cot (x-b)-\cot (x-a)) d x \\
&=\frac{1}{\sin (b-a)}(\log |\sin (x-b)|-\log |\sin n(x-a)|)+C
\end{aligned}
$$
$=\operatorname{cosec}(b-a) \log \left|\frac{\sin (x-b)}{\sin (x-a)}\right|+C$
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