We are given, I=∫sin (x α)sin (x+a)dx I=∫sin (x+α 2α)sin (x+α)dx [ Now we know that sin (A B)=sin Acos B cos Asin B ] I=∫[sin (x+α)cos (2α) c ...

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Implicit Differentiation

∫sin x-αsin x...

Question

$\int \frac{sin (x - α)}{sin (x + α)} d x$ .

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Solution

We are given, $I = \int \frac{sin (x - α)}{sin (x + a)} d x$
$I = \int \frac{sin (x + α - 2 α)}{sin (x + α)} d x$
[ Now we know that $sin (A - B) = sin A cos B - cos A sin B$ ]
$I = \int \frac{[sin (x + α) cos (2 α) - cos (x α) sin (2 α)]}{sin (x + α)} d x$
$I = \int cos 2 α d x - i n t n cot (x + α) . sin 2 α d x$
$I = cos 2 α . x - ln | sin (x + α) | . sin 2 α$
$I = x . cos 2 α - sin 2 α . ln | sin (x + α) | + C$

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