A) \[\pi /12\]
B) \[\pi /4\]
C) \[\pi /3\]
D) \[\pi /6\]
E) \[\pi /2\]
Correct Answer: A
Solution :
Let \[I=\int_{\pi /6}^{\pi /3}{\frac{1}{1+{{\tan }^{3}}x}}dx\] \[\Rightarrow \] \[I=\int_{\pi /6}^{\pi /3}{\frac{{{\cos }^{3}}x}{{{\sin }^{3}}x+{{\cos }^{3}}x}}dx\] ?.(i) \[\Rightarrow \] \[I=\int_{\pi /6}^{\pi /3}{\frac{{{\cos }^{3}}\left( \frac{\pi }{2}-x \right)}{{{\sin }^{3}}\left( \frac{\pi }{2}-x \right)+{{\cos }^{3}}\left( \frac{\pi }{2}-x \right)}}dx\] \[\Rightarrow \] \[I=\int_{\pi /6}^{\pi /3}{\frac{{{\sin }^{3}}x}{{{\sin }^{3}}x+{{\cos }^{3}}x}}dx\] ?. (ii) On adding Eqs. (i) and (ii), we get \[2I=\int_{\pi /6}^{\pi /6}{\frac{{{\sin }^{3}}x+{{\cos }^{3}}x}{{{\sin }^{3}}x+{{\cos }^{3}}x}}dx\] \[=\int_{\pi /6}^{\pi /3}{1\,dx}=[x]_{\pi /6}^{\pi /3}\] \[=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}\] \[\Rightarrow \] \[I=\frac{\pi }{12}\]
You need to login to perform this action.
You will be redirected in
3 sec
