Definite Integration Question 55
Question: The value of $ \int _{0}^{\pi /2}{\frac{{{\sin }^{2/3}}x}{{{\sin }^{2/3}}x+{{\cos }^{2/3}}x}dx} $ is
[RPET 2001]
Options:
A) $ \pi /4 $
B) $ \pi /2 $
C) $ 3\pi /4 $
D) $ \pi $
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_0^{\pi /2}{\frac{{{\sin }^{2/3}}x}{{{\sin }^{2/3}}x+{{\cos }^{2/3}}x}dx} $
or $ I=\int_0^{\pi /2}{\frac{{{\sin }^{2/3}}( \frac{\pi }{2}-x )}{{{\sin }^{2/3}}( \frac{\pi }{2}-x )+{{\cos }^{2/3}}( \frac{\pi }{2}-x )}dx} $
or $ I=\int_0^{\pi /2}{\frac{{{\cos }^{2/3}}x}{{{\cos }^{2/3}}x+{{\sin }^{2/3}}x}}dx $
Therefore, $ 2I=\int\limits_0^{\pi /2}{\frac{({{\sin }^{2/3}}x+{{\cos }^{2/3}}x)}{({{\sin }^{2/3}}x+{{\cos }^{2/3}}x)}dx} $
$ \Rightarrow 2I=\int _{0}^{\pi /2}{dx} $
$ \Rightarrow I=\frac{1}{2}[x]_0^{\pi /2} $
$ =\frac{\pi }{4} $ .
Trick: $ \int_0^{\pi /2}{\frac{{{\sin }^{n}}x}{{{\sin }^{n}}x+{{\cos }^{n}}x}}dx=\frac{\pi }{4} $ .
BETA
