Definite Integration Question 382
Question: $ \int_0^{\pi /2}{\frac{\sin x\cos xdx}{{{\cos }^{2}}x+3\cos x+2}}= $
[MNR 1981]
Options:
A) $ \log ( \frac{8}{9} ) $
B) $ \log ( \frac{9}{8} ) $
C) $ \log (8\times 9) $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ I=\int_0^{\pi /2}{\frac{\sin x\cos x.dx}{{{\cos }^{2}}x+3\cos x+2}} $
We put $ \cos x=t\Rightarrow -\sin xdx=dt, $ then $ I=\int_0^{1}{\frac{t.dt}{t^{2}+3t+2}=\int_0^{1}{[ \frac{2}{t+2}-\frac{1}{t+1} ]}}dt $
$ =[2\log (t+2)-\log (t+1)]_0^{1} $
$ =[2\log 3-\log 2-2\log 2] $
$ =[2\log 3-3\log 2]=[\log 9-\log 8]=\log ( \frac{9}{8} ) $ .
BETA
