sin-x-sin-x-a-dx-

Question Number 151221 by peter frank last updated on 19/Aug/21

\int \frac{\sin x}{\sin (x - a)} dx

Commented by puissant last updated on 19/Aug/21

Q = \int \frac{s i n x}{s i n (x - a)} d x

= \int \frac{s i n (x - a + a)}{s i n (x - a)} d x

= \int \frac{s i n (x - a) c o s a + c o s (x - a) s i n a}{s i n (x - a)} d x

= \int {c o s a + c o t a n (x - a) s i n a} d x

∵∴ Q = x c o s a + l n ∣ s i n (x - a) ∣ s i n a + C . .

Commented by peter frank last updated on 19/Aug/21

thank you

Answered by Olaf_Thorendsen last updated on 19/Aug/21

Let F_{b} (x) = \int_{b}^{x} \frac{\sin t}{\sin (t - a)} d t

F_{b} (x) = \int_{b - a}^{x - a} \frac{\sin (u + a)}{\sin u} d u

F_{b} (x) = \int_{b - a}^{x - a} \frac{\sin u \cos a + \cos u \sin a}{\sin u} d u

F_{b} (x) = {[u \cos a + \ln ∣ \sin u ∣ . \sin a]}_{b - a}^{x - a}

F_{b} (x) = (x - b) \cos a + \ln ∣ \frac{\sin (x - a)}{\sin (b - a)} ∣ . \sin a

F (x) = x \cos a + \ln ∣ \sin (x - a) ∣ + C