Chapter 7.3, Problem 2QQ

To determine

To find which of the following integral is correct.

Answer to Problem 2QQ

  $\text{The correct answer is C}\text{.}$

Explanation of Solution

Given information:

  $\begin{array}{l}\text{The integrand of }{\displaystyle {\int }_0^{1/2}\frac{\sqrt{x}}{\sqrt{1-x}}dx},\text{ where }\sqrt{x}=\sin y\\ \left(\text{A}\right){\displaystyle {\int }_0^{1/2}{\sin }^{2}y}dy\left(\text{B}\right)\text{ 2 }{\displaystyle {\int }_0^{1/2}\frac{{\sin }^{2}y}{\cos y}}dy\\ \left(C\right)\text{ 2 }{\displaystyle {\int }_0^{\pi /4}{\sin }^{2}y}dy\left(\text{D}\right){\displaystyle {\int }_0^{\pi /4}{\sin }^{2}y}dy\\ \left(\text{E}\right)\text{ 2 }{\displaystyle {\int }_0^{\pi /6}{\sin }^{2}y}dy\end{array}$

Calculation : The required solution is obtained as,

  $\begin{array}{l}\sqrt{x}=\sin y\\ x={\sin }^{2}y\left(\text{squaring both side}\right)\\ dx=2\sin y\cos ydy\left(\text{differentiate using chain rule}\right)\\ \text{ Plugging in the original equation}\text{.}\\ \frac{\sin y}{\sqrt{1-{\sin }^{2}y}}\\ \text{Calculate the bounds for the integral since we}\\ \text{have substituted in }y\text{ for }x.\\ \frac{1}{2}={\sin }^{2}y\\ \frac{\sqrt{2}}{2}=\sin y\\ y=\frac{\pi }{4}\\ \text{ 0}={\sin }^{2}y\\ y=0\\ \\ \text{Use the new bounds we calculated in the last step,}\\ \text{then plug in dx from the first step}\text{.}\\ {\displaystyle {\int }_0^{\pi /4}\frac{\sin y}{\sqrt{1-{\sin }^{2y}}}}2\sin y\cos ydy\\ {\displaystyle {\int }_0^{\pi /4}\frac{\sin y}{\cos y}}\times 2\sin y\cos ydy\left(1-{\sin }^{2}y={\cos }^{2}y\right)\\ \text{ 2}{\displaystyle {\int }_0^{\pi /4}{\sin }^2}ydy\\ \\ \text{Hence, the correct answer is C}\text{.}\end{array}$