A) \[\pi /4\]
B) \[\pi /2\]
C) \[3\pi /2\]
D) \[2\pi \]
E) \[\pi \]
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) \[I=4\int_{0}^{\pi /2}{\frac{{{\sin }^{4}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{4}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{4}}\left( \frac{\pi }{2}-x \right)}\ dx}\] \[I=4\int_{0}^{\pi /2}{\frac{{{\cos }^{4}}x}{{{\cos }^{4}}x+{{\sin }^{4}}x}\ dx}\] .....(ii) Adding (i) and (ii) we get,\[2I=4\int_{0}^{\pi /2}{dx=4\times \frac{\pi }{2}=2\pi }\] Þ \[I=\pi \].
You need to login to perform this action.
You will be redirected in
3 sec
