Chapter 7.2, Problem 83E

(a)

To determine

To show ${\displaystyle \underset{0}{\overset{1/2}{\int }}\frac{\sqrt{x}dx}{\sqrt{1-x}}}={\displaystyle \underset{0}{\overset{\pi /4}{\int }}2{\sin }^{2}ydy}$ by substitution.

Explanation of Solution

Given information: $\sqrt{x}=\sin y$

Calculation :

Put $x={\sin }^{2}y$ so that $dx=2\sin y\cos ydy$

Also, when $x=0;y=0$ and when $x=1/2;y=\pi /4$

The given integral becomes as:

  ${\displaystyle \underset{0}{\overset{1/2}{\int }}\frac{\sqrt{x}dx}{\sqrt{1-x}}}={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\frac{\sin y(2\sin y\cos y)}{\sqrt{1-{\sin }^{2}y}}dy}={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\frac{2{\sin }^{2}y\cos y}{\cos y}dy}={\displaystyle \underset{0}{\overset{\pi /4}{\int }}2{\sin }^{2}ydy}$

This is the required proof

(b)

To determine

To evaluate ${\displaystyle \underset{0}{\overset{1/2}{\int }}\frac{\sqrt{x}dx}{\sqrt{1-x}}}$

Explanation of Solution

Given information: $\sqrt{x}=\sin y$

Calculation :

  ${\displaystyle \underset{0}{\overset{1/2}{\int }}\frac{\sqrt{x}dx}{\sqrt{1-x}}}={\displaystyle \underset{0}{\overset{\pi /4}{\int }}2{\sin }^{2}ydy}={\displaystyle \underset{0}{\overset{\pi /4}{\int }}2\left(\frac{1-\cos 2y}{2}\right)dy}$

  $={\displaystyle \underset{0}{\overset{\pi /4}{\int }}\left(1-\cos 2y\right)dy}={\left[y-\frac{\sin 2y}{2}\right]}_0^{\pi /4}=\left[\frac{\pi }{4}-\frac{1}{2}\right]$

The value of the integral is $\left[\frac{\pi }{4}-\frac{1}{2}\right]$