Definite Integration Question 362
Question: $ \int_0^{\pi /4}{\frac{\sin x+\cos x}{9+16\sin 2x}dx=} $
[IIT 1983]
Options:
A) $ \frac{1}{20}\log 3 $
B) $ \log 3 $
C) $ \frac{1}{20}\log 5 $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ I=\int_0^{\pi /4}{\frac{\sin x+\cos x}{9+16\sin 2x}}dx $
Put $ \sin x-\cos x=t $ , then $ (\sin x+\cos x)dx=dt $
$ I=\int _{-1}^{0}{\frac{dt}{9+16(1-t^{2})}}=\int _{-1}^{0}{\frac{dt}{25-16t^{2}}} $
$ =\frac{1}{10}\int _{-1}^{0}{( \frac{1}{5-4t}+\frac{1}{5+4t} )dt} $
$ =| \frac{1}{10}.\frac{1}{4}[\log (5+4t)-\log (5-4t)] | _{-1}^{0} $
$ =\frac{1}{40}(\log 9-\log 1)=\frac{1}{20}\log 3 $ .
BETA
