A) \[\pi /4\]
B) \[\pi /2\]
C) \[3\pi /4\]
D) \[\pi \]
Correct Answer: A
Solution :
\[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{2/3}}x}{{{\sin }^{2/3}}x+{{\cos }^{2/3}}x}dx}\] or \[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{2/3}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{2/3}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{2/3}}\left( \frac{\pi }{2}-x \right)}dx}\] 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: \[\int_{0}^{\pi /2}{\frac{{{\sin }^{n}}x}{{{\sin }^{n}}x+{{\cos }^{n}}x}\,}dx=\frac{\pi }{4}\].
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