Definite Integration Question 395
Question: $ \int_0^{\pi }{\frac{dx}{1-2a\cos x+a^{2}}} $ =
[CEE 1993]
Options:
A) $ \frac{\pi }{2(1-a^{2})} $
B) $ \pi (1-a^{2}) $
C) $ \frac{\pi }{1-a^{2}} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_0^{\pi }{\frac{dx}{(1+a^{2})( {{\cos }^{2}}\frac{x}{2}+{{\sin }^{2}}\frac{x}{2} )-2a( {{\cos }^{2}}\frac{x}{2}-{{\sin }^{2}}\frac{x}{2} )}} $
$ =\int_0^{\pi }{\frac{dx}{{{(1-a)}^{2}}{{\cos }^{2}}\frac{x}{2}+{{(1+a)}^{2}}{{\sin }^{2}}\frac{x}{2}}} $
$ =\frac{2}{{{(1+a)}^{2}}}\int_0^{\infty }{\frac{dt}{{{{ (1-a)/(1+a) }}^{2}}+t^{2}}} $ ; {where $ t=\tan \frac{x}{2} $ } $ =\frac{2}{{{(1+a)}^{2}}}\frac{(1+a)}{(1-a)}[ {{\tan }^{-1}}( \frac{1+a}{1-a}.t ) ]_0^{\infty } $
$ =\frac{2}{(1-a^{2})}[{{\tan }^{-1}}\infty -{{\tan }^{-1}}0]=\frac{\pi }{1-a^{2}} $ .
BETA
