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