Definite Integration Question 394
Question: $ \int_0^{\pi /2}{\frac{1+2\cos x}{{{(2+\cos x)}^{2}}}=} $
[CEE 1993]
Options:
A) $ \frac{\pi }{2} $
B) $ \pi $
C) $ \frac{1}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_0^{\pi /2}{\frac{(1+2\cos x)}{{{(2+\cos x)}^{2}}}dx=\int_0^{\pi /2}{\frac{2(\cos x+2)-3}{{{(2+\cos x)}^{2}}}dx}} $
$ =2\int_0^{\pi /2}{\frac{dx}{2+\cos x}-3\int_0^{\pi /2}{\frac{dx}{{{(2+\cos x)}^{2}}}}} $
$ =4\int_0^{1}{\frac{dt}{3+t^{2}}-6\int_0^{1}{\frac{1+t^{2}}{{{(3+t^{2})}^{2}}}dt}} $ , $ [ Put\tan \frac{x}{2}=t ] $
$ =-2\int_0^{1}{\frac{dt}{3+t^{2}}+12\int_0^{1}{\frac{dt}{{{(3+t^{2})}^{2}}}}} $
$ =-2\int_0^{1}{\frac{dt}{3+t^{2}}+12}[ \frac{1}{6}.\frac{t}{t^{2}+3} ]_0^{1}+\frac{1}{6}\int_0^{1}{\frac{dt}{3+t^{2}}} $
$ =2[ \frac{t}{t^{2}+3} ]_0^{1}=\frac{1}{2} $ .
BETA
