_ ^ x (x - ) dx = - Vidyadip

MCQ

$\int_{}^{} {\frac{{\sin x}}{{\sin (x - \alpha )}}dx = } $

A
$x\cos \alpha - \sin \alpha \log \sin (x - \alpha ) + c$
✓
$x\cos \alpha + \sin \alpha \log \sin (x - \alpha ) + c$
C
$x\sin \alpha - \sin \alpha \log \sin (x - \alpha ) + c$
D
None of these

✓

Answer

Correct option: B.

$x\cos \alpha + \sin \alpha \log \sin (x - \alpha ) + c$

b
(b)$\int_{}^{} {\frac{{\sin x}}{{\sin (x - \alpha )}}\,dx = \int_{}^{} {\frac{{\sin (x - \alpha + \alpha )}}{{\sin (x - \alpha )}}\,dx} } $
$ = \int_{}^{} {\frac{{\left\{ {(\sin (x - \alpha )\cos \alpha + \cos (x - \alpha )\sin \alpha } \right\}}}{{\sin (x - \alpha )}}\,dx} $
$ = \int_{}^{} {\cos \alpha \,dx + \int_{}^{} {\sin \alpha \,.\,\cot \,(x - \alpha )\,dx} } $
$ = x\cos \alpha + \sin \alpha \,.\,\log \sin (x - \alpha ) + c$.

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