Chapter 6.3, Problem 2QQ

To determine

To rewrite: the given integral by using the substitution.

Answer to Problem 2QQ

Option C is correct.

Explanation of Solution

Given:

  ${\displaystyle \underset{0}{\overset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{\sqrt{x}}{\sqrt{1-x}}dx}$ .

Concept used:

The Fundamental Theorem of Calculus, Evaluation Part:

If $f$ is continuous at every point of $\left[a,b\right]$ , and if $F$ is any antiderivative of $f$ on $\left[a,b\right]$ , then

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=F\left(b\right)-F\left(a\right)$ .

Calculation:

Consider the given integral,

  ${\displaystyle \underset{0}{\overset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{\sqrt{x}}{\sqrt{1-x}}dx}$

Put $\sqrt{x}=\sin y$ . Then,

  $x={\sin }^{2}y$

Now by differentiating with respect to $x$ , it follows that

  $\begin{array}{c}1=2\sin y\cos y\frac{dy}{dx}\\ dx=2\sin y\cos ydy\end{array}$

Now when $x=0$ ,

  $\begin{array}{c}\sin y=0\\ y={\sin }^{-1}\left(0\right)\\ y=0\end{array}$

And when $x=\frac{1}{2}$ ,

  $\begin{array}{c}\sin y=\frac{1}{2}\\ y={\sin }^{-1}\left(\frac{1}{2}\right)\\ y=\frac{\pi }{6}\end{array}$

Therefore the integral becomes,

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\frac{\sqrt{x}}{\sqrt{1-x}}dx}={\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}\frac{\sin y}{\sqrt{1-{\sin }^{2}y}}2\sin y\cos ydy}\\ =2{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}\frac{{\sin }^{2}y\cos y}{\sqrt{{\cos }^{2}y}}dy}\\ =2{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}{\sin }^{2}ydy}\end{array}$

Hence option C is correct.