Definite Integration Question 323
Question: $ \int_0^{2\pi }{\frac{\sin 2\theta }{a-b\cos \theta }d\theta =} $
[Roorkee 1988]
Options:
A) 1
B) 2
C) $ \frac{\pi }{4} $
D) 0
Show Answer
Answer:
Correct Answer: D
Solution:
$ I=\int_0^{2\pi }{\frac{\sin 2\theta }{a-b\cos \theta }d\theta =\int_0^{2\pi }{\frac{\sin (2\pi -2\theta )}{a-b\cos (2\pi -\theta )}d\theta }} $
therefore I $ =-\int_0^{2\pi }{\frac{\sin 2\theta }{a-b\cos \theta }d\theta } $
$ \Rightarrow 2I=0\Rightarrow \int_0^{2\pi }{\frac{\sin 2\theta }{a-b\cos \theta }d\theta =0} $ .
BETA
