Mathematics | VITEEE 2019 | Discussion Page

If $\int_{}^{}\frac{\sin x}{\sin (x - \alpha)}dx = Ax + B\log_{}{\sin (x - \alpha)}+C$ then value of (A,B) is

$(-\cos\alpha,\sin\alpha)$

$(\cos\alpha,\sin\alpha)$

$(-\sin\alpha,\cos\alpha)$

$(\sin\alpha,\cos\alpha)$

Option B

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

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

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

= $\int_{}^{}\left\{\cos\alpha+\sin\alpha\cot(x - \alpha)\right\}dx$

= $(\cos\alpha)x+(\sin\alpha)\log_{}{\sin(x-\alpha)}+C$

A = $\cos\alpha$ B = $\sin\alpha$

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