Definite Integration Question 417
Question: $ \int_0^{\pi /4}{\frac{\sec x}{1+2{{\sin }^{2}}x}} $ is equal to
[MNR 1994]
Options:
A) $ \frac{1}{3}[ \log (\sqrt{2}+1)+\frac{\pi }{2\sqrt{2}} ] $
B) $ \frac{1}{3}[ \log (\sqrt{2}+1)-\frac{\pi }{2\sqrt{2}} ] $
C) $ 3[ \log (\sqrt{2}+1)-\frac{\pi }{2\sqrt{2}} ] $
D) $ 3[ \log (\sqrt{2}+1)+\frac{\pi }{2\sqrt{2}} ] $
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ I=\int_0^{\pi /4}{\frac{\cos x}{{{\cos }^{2}}x(1+2{{\sin }^{2}}x)}}dx $
$ =\int_0^{\pi /4}{\frac{\cos xdx}{(1-{{\sin }^{2}}x)(1+2{{\sin }^{2}}x)}} $
$ =\frac{1}{3}\int_0^{1/\sqrt{2}}{( \frac{1}{1-t^{2}}+\frac{2}{1+2t^{2}} )}dt $
By partial fractions, where $ t=\sin x $
$ =\frac{1}{3}[ \frac{1}{2.1}\log \frac{1+t}{1-t}+\frac{2}{\sqrt{2}}{{\tan }^{-1}}t\sqrt{2} ]_0^{1/\sqrt{2}} $
$ =\frac{1}{3}[ \frac{1}{2}\log \frac{(\sqrt{2}+1)}{(\sqrt{2}-1)}+\sqrt{2}{{\tan }^{-1}}1 ] $
$ =\frac{1}{3}[ \frac{1}{2}\log {{(\sqrt{2}+1)}^{2}}+\sqrt{2}.\frac{\pi }{4} ] $
$ =\frac{1}{3}[ \log (\sqrt{2}+1)+\frac{\pi }{2\sqrt{2}} ] $ .
BETA
