If $\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+C$, then value of $(A, B)$ is

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

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

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

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

JEE Main · 2004 · 24 Apr 2004 · Offline · Indefinite Integrals

\int \frac{s i n x}{s i n ( x - α )} d x = A x + B lo g sin (x - α) + C

, then value of

(A, B)

Answer: (B)

Write the numerator as

sin x = sin [(x - α) + α] = sin (x - α) cos α + cos (x - α) sin α

. Dividing by

sin (x - α)

gives

\frac{s i n x}{s i n ( x - α )} = cos α + sin α cot (x - α)

. Integrating, the first term yields

x cos α

and the second yields

sin α lo g ∣ sin (x - α) ∣

, so matching against

A x + B lo g sin (x - α) + C

gives

A = cos α

and

B = sin α

JEE Main · 2004 · 24 Apr 2004 · Offline · Single correct option