A) 0
B) \[\pi \]
C) \[\pi /2\]
D) \[\pi /4\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{3/2}}x\,dx}{{{\cos }^{3/2}}x+{{\sin }^{3/2}}x}}\] ?..(i) = \[\int_{0}^{\pi /2}{\frac{{{\sin }^{3/2}}\left( \frac{\pi }{2}-x \right)}{{{\cos }^{3/2}}\left( \frac{\pi }{2}-x \right)+{{\sin }^{3/2}}\left( \frac{\pi }{2}-x \right)}dx}\] = \[\int_{0}^{\pi /2}{\frac{{{\cos }^{3/2}}x\,dx}{{{\sin }^{3/2}}x+{{\cos }^{3/2}}x}}\] .....(ii) Adding (i) and (ii), we get \[I=\frac{1}{2}\int_{0}^{\pi /2}{1dx=\frac{1}{2}[x]_{0}^{\pi /2}=\frac{\pi }{4}}\].
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