MCQ
$\int_{\,0}^{\,\pi /2} {\frac{{{{\sin }^{2/3}}x}}{{{{\sin }^{2/3}}x + {{\cos }^{2/3}}x}}dx} =$
- ✓$\pi /4$
- B$\pi /2$
- C$3\pi /4$
- D$\pi $
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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