The valueof the definite integral \(\int_{0}^{\frac{π}{2}}\)\(\sqrt {tan  x}\) dx is

\(\frac{π}{\sqrt{2}}\)

The correct answer is option 3. \(\frac{π}{\sqrt{2}}\).

\(I=\int\limits_{0}^{\frac{π}{2}}\sqrt {tan  x}dx\) ....(i)

Using property,

\(I=\int\limits_{0}^{\frac{π}{2}}\sqrt {cot  x}dx\) ....(ii)

Adding eq. (i) & (ii)

\(2I=\int\limits_{0}^{\frac{π}{2}}(\sqrt {tan  x}+\sqrt {cot  x})dx\)

\(⇒ I = \frac{1}{\sqrt{2}}\int\limits_{0}^{\frac{\pi }{2}}\frac{sin x + cos x}{\sqrt{sin 2x}}dx\)

\(⇒ I = \frac{1}{\sqrt{2}}\int\limits_{0}^{\frac{\pi }{2}}\frac{sin x + cos x}{\sqrt{1 - (sin x - cos x)^2}}dx\)

Let,

\(sinx - cosx = t\)

\(⇒ (cos x + sin x)dx = dt\)

\(∴ I = \frac{1}{\sqrt{2}}\int\limits_{-1}^{1}\frac{dt}{\sqrt{1 - t^2}}\)

\(⇒ I = \frac{1}{\sqrt{2}}[sin^{-1} t]^1_{-1} = \frac{\pi }{2}\)