question_answer

Using properties of definite integrals,  evaluate \[\int_{\pi /6}^{\pi /3}{\frac{dx}{1+\sqrt{\tan \,\,x}}.}\]

Answer:

Let \[l=\int_{\pi /6}^{\pi /3}{\frac{dx}{1+\sqrt{\tan \,x}}=\int_{\pi /6}^{\pi /3}{\frac{dx}{1+\frac{\sqrt{\sin \,x}}{\sqrt{\cos \,x}}}}}\] \[\Rightarrow \]   \[l=\int_{\pi /6}^{\pi /3}{\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx}\]         ?(i) \[\Rightarrow \] \[l=\int_{\pi /6}^{\pi /3}{\frac{\sqrt{\cos \left( \frac{\pi }{6}+\frac{\pi }{3}-x \right)}}{\sqrt{\cos \left( \frac{\pi }{6}+\frac{\pi }{3}-x \right)}+\sqrt{\sin \left( \frac{\pi }{6}+\frac{\pi }{3}-x \right)}}dx}\]                          \[\left[ \because \,\,\int_{a}^{b}{f(x)dx=\int_{a}^{b}{f(a+b-x)dx}} \right]\] \[\Rightarrow \] \[l=\int_{\pi /6}^{\pi /3}{\frac{\sqrt{\cos \left( \frac{\pi }{2}-x \right)}}{\sqrt{\cos \left( \frac{\pi }{2}-x \right)}+\sqrt{\sin \left( \frac{\pi }{2}-x \right)}}\,dx}\] \[\Rightarrow \] \[l=\int_{\pi /6}^{\pi /3}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx}\]                   ?(ii) On adding Eqs. (i) and (ii), we get \[2l=\int_{\pi /6}^{\pi /3}{\frac{\sqrt{\cos x}+\sqrt{\sin x}}{\sqrt{\cos x}+\sqrt{sinx}}\,dx}\] \[\Rightarrow \]   \[2l=\int_{\pi /6}^{\pi /3}{1\,dx\,=[x]_{\pi /6}^{\pi /3}=\frac{\pi }{3}-\frac{\pi }{6}}\] \[\Rightarrow \]   \[2l=\frac{2\pi -\pi }{6}=\frac{\pi }{6}\] \[\Rightarrow \]   \[l=\frac{\pi }{12}\]