Question

Integrate ${\int }_0^{\frac{\pi }{2}}\sqrt{\sin \left(x\right)}dx$

Answer 1

Step 1. Evaluate the integral ${\int }_0^1\sqrt{1-z^2}dz$

Let $I_1={\int }_0^1\sqrt{1-z^2}dz$

Using the method of Integration by parts we get

$$\begin{gathered} \Rightarrow I_1={\left[z\sqrt{1-z^2}\right]}_0^1-{\int }_0^1\left[\frac{d\sqrt{1-z^2}}{dz}\int dz\right]dz \\ \Rightarrow I_1=0-{\int }_0^1\left[\frac{-z}{\sqrt{1-z^2}}z\right]dz \\ \Rightarrow I_1={\int }_0^1\frac{z^2-1+1}{\sqrt{1-z^2}}dz \\ \Rightarrow I_1=-{\int }_0^1\sqrt{1-z^2}+{\int }_0^1\frac{1}{\sqrt{1-z^2}}dz \\ \Rightarrow I_1=-I_1+{\left[{\sin }^{-1}\left(z\right)\right]}_0^1 \\ \Rightarrow 2I_1=\frac{\pi }{2} \\ \Rightarrow I_1=\frac{\pi }{4}.....................\left(i\right) \end{gathered}$$

Step 2. Find the value of the given definite integral

Given $I=$${\int }_0^{\frac{\pi }{2}}\sqrt{\sin \left(x\right)}dx$

Let $\sqrt{\sin \left(x\right)}=z$

$$\begin{gathered} \therefore dz=\frac{\cos \left(x\right)}{2\sqrt{\sin \left(x\right)}}dx \\ \Rightarrow \frac{2z}{\sqrt{1-z^2}}dz=dx \end{gathered}$$

$$\begin{gathered} I={\int }_0^1\frac{2z^2}{\sqrt{1-z^2}}dz \\ =2{\int }_0^1\frac{z^2-1+1}{\sqrt{1-z^2}}dz \\ =2{\int }_0^1\frac{1}{\sqrt{1-z^2}}dz-2{\int }_0^1\frac{1-z^2}{\sqrt{1-z^2}}dz \\ =2{\left[{\sin }^{-1}\left(z\right)\right]}_0^1-2{\int }_0^1\sqrt{1-z^2}dz \\ =2\left[{\sin }^{-1}\left(1\right)-{\sin }^{-1}\left(0\right)\right]-2{\int }_0^1\sqrt{1-z^2}dz \\ =2\left[\frac{\pi }{2}-0\right]-2{\int }_0^1\sqrt{1-z^2}dz \\ =\pi -2{\int }_0^1\sqrt{1-z^2}dz \\ =\pi -2{\int }_0^1\sqrt{1-z^2}dz \\ =\pi -2\left(\frac{\pi }{4}\right)...\left[\mathrm{From}\left(i\right)\right] \\ =\frac{\pi }{2} \end{gathered}$$