Definite Integration Question 126
Question: $ \int _{\ -\pi }^{\pi }{\frac{{{\sin }^{4}}x}{{{\sin }^{4}}x+{{\cos }^{4}}x}\ dx}= $
[Kerala (Engg.) 2005]
Options:
A) $ \pi /4 $
B) $ \pi /2 $
C) $ 3\pi /2 $
D) $ 2\pi $
E) $ \pi $
Show Answer
Answer:
Correct Answer: E
Solution:
$ I=\int _{-\pi }^{\pi }{\frac{{{\sin }^{4}}x}{{{\sin }^{4}}x+{{\cos }^{4}}x}\ dx} $
$ \therefore $ $ I=2\times 2\int_0^{\pi /2}{\frac{{{\sin }^{4}}x}{{{\sin }^{4}}x+{{\cos }^{4}}x}\ dx} $
…..(i) $ I=4\int_0^{\pi /2}{\frac{{{\sin }^{4}}( \frac{\pi }{2}-x )}{{{\sin }^{4}}( \frac{\pi }{2}-x )+{{\cos }^{4}}( \frac{\pi }{2}-x )}\ dx} $
$ I=4\int_0^{\pi /2}{\frac{{{\cos }^{4}}x}{{{\cos }^{4}}x+{{\sin }^{4}}x}\ dx} $ …..(ii)
Adding (i) and (ii) we get,
$ 2I=4\int_0^{\pi /2}{dx=4\times \frac{\pi }{2}=2\pi } $
therefore $ I=\pi $ .
BETA
