A) \[\pi /2\]
B) \[\pi /4\]
C) \[\pi \]
D) None of these
Correct Answer: B
Solution :
Let \[I=\int_{0}^{\pi /2}{\frac{1}{1+{{\tan }^{3}}x}}\,dx\] \[\Rightarrow \] \[I=\int_{0}^{\pi /2}{\frac{{{\cos }^{3}}x}{{{\cos }^{3}}\,x+{{\sin }^{3}}x}}dx\] ?..(i) \[\Rightarrow \] \[I=\int_{0}^{\pi /2}{\frac{{{\cos }^{3}}\left( \frac{\pi }{2}-x \right)}{{{\cos }^{3}}\left( \frac{\pi }{2}-x \right)+{{\sin }^{3}}\left( \frac{\pi }{2}-x \right)}}dx\] \[\Rightarrow \] \[I=\int_{0}^{\pi /2}{\frac{{{\sin }^{3}}x}{{{\sin }^{3}}x+{{\cos }^{3}}x}dx}\] ?.(ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{0}^{\pi /2}{1\,\,\,dx}\] \[\Rightarrow \] \[2I=[x]_{0}^{\pi /2}\] \[\Rightarrow \] \[I=\frac{\pi }{4}\]
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