Integral Calculus> evaluate the following integral ∫ sin 2x ...

evaluate the following integral
∫ sin 2x / sin(x – pi/6) sin(x + pi/6) dx

taniska , 11 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 11 Years ago

Ans:
Hello student,
Please find the answer to your question below

$I = \int \frac{sin2x}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx$
$I = \int \frac{sin(x+\frac{\pi }{6}+x-\frac{\pi }{6})}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx$
$I = \int \frac{sin(x+\frac{\pi }{6})cos(x-\frac{\pi }{6})+cos(x+\frac{\pi }{6})sin(x-\frac{\pi }{6})}{sin(x-\frac{\pi }{6})sin(x+\frac{\pi }{6})}dx$
$I = \int \frac{cos(x-\frac{\pi }{6})}{sin(x-\frac{\pi }{6})}dx+\int \frac{cos(x+\frac{\pi }{6})}{sin(x+\frac{\pi }{6})}dx$
$I = log(sin(x-\frac{\pi }{6}))+log(sin(x+\frac{\pi }{6}))+c$