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