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