JEE MAIN - Mathematics (2004 - No. 16)
If $$\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}dx = Ax + B\log \sin \left( {x - \alpha } \right), + C,} $$ then value of
$$(A, B)$$ is
$$(A, B)$$ is
$$\left( { - \cos \alpha ,\sin \alpha } \right)$$
$$\left( { \cos \alpha ,\sin \alpha } \right)$$
$$\left( { - \sin \alpha ,\cos \alpha } \right)$$
$$\left( { \sin \alpha ,\cos \alpha } \right)$$
Explanation
$$\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}} dx$$
$$ = \int {{{\sin \left( {x - \alpha + \alpha } \right)} \over {\sin \left( {x - \alpha } \right)}}} dx$$
$$ = \int {{{\sin \left( {x - \alpha } \right)\cos \alpha + \cos \left( {x - \alpha } \right)\sin \alpha } \over {\sin \left( {x - \alpha } \right)}}} $$
$$ = \int {\left\{ {\cos \alpha + \sin \alpha \,\cot \left. {\left( {x - \alpha } \right)} \right\}} \right.} dx$$
$$ = \left( {\cos \alpha } \right)x + \left( {\sin \alpha } \right)\log \,\sin \left( {x - \alpha } \right) + C$$
$$\therefore$$ $$A = \cos \alpha ,$$ $$B = \sin \alpha $$
$$ = \int {{{\sin \left( {x - \alpha + \alpha } \right)} \over {\sin \left( {x - \alpha } \right)}}} dx$$
$$ = \int {{{\sin \left( {x - \alpha } \right)\cos \alpha + \cos \left( {x - \alpha } \right)\sin \alpha } \over {\sin \left( {x - \alpha } \right)}}} $$
$$ = \int {\left\{ {\cos \alpha + \sin \alpha \,\cot \left. {\left( {x - \alpha } \right)} \right\}} \right.} dx$$
$$ = \left( {\cos \alpha } \right)x + \left( {\sin \alpha } \right)\log \,\sin \left( {x - \alpha } \right) + C$$
$$\therefore$$ $$A = \cos \alpha ,$$ $$B = \sin \alpha $$
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