Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 41 Maths Textbook Solution.

$\frac{\pi }{8}$

Hint: Use indefinite integral formula then put limits to solve this integral

Given:

$\int_{0}^{1}\sqrt{x\left (1-x \right )}dx$

Solution:

$\int_{0}^{1}\sqrt{x\left (1-x \right )}dx$

Put

$x=\sin ^{2}\theta \Rightarrow dx=2\sin \theta .\cos \theta d\theta$

When $x=0$ then

$\theta =0$

When $x=1$ then

$\theta =\frac{\pi }{2}$

Then

$\int_{0}^{1} \sqrt{x(1-x)} d x=\int_{0}^{\frac{\pi}{2}} \sqrt{\sin ^{2} \theta\left(1-\sin ^{2} \theta\right)} \cdot 2 \sin \theta \cdot \cos \theta d \theta$

$\left[\because 1-\sin ^{2} \theta=\cos ^{2} \theta\right]$

$=\int_{0}^{\frac{\pi}{2}} \sin \theta \sqrt{\cos ^{2} \theta} \cdot 2 \sin \theta \cdot \cos \theta d \theta$

$=\int_{0}^{\frac{\pi}{2}} \sin \theta \cos \theta \cdot 2 \sin \theta \cdot \cos \theta d \theta$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} 4 \sin ^{2} \theta \cdot \cos ^{2} \theta d \theta$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}(2 \sin \theta \cdot \cos \theta)^{2} d \theta$

$\left [ \because \sin 2\theta =2\sin \theta \cos \theta \right ]$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}(\sin 2 \theta)^{2} d \theta$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \sin ^{2} 2 \theta d \theta$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(\frac{1-\cos 2 \theta}{2}\right) d \theta$

$\left[\begin{array}{l}\because 2 \sin ^{2} \theta=1-\cos 2 \theta \\ \Rightarrow \sin ^{2} \theta=\frac{1-\cos 2 \theta}{2}\end{array}\right]$

$=\frac{1}{2} \int_{0}^{\frac{\pi}{2}}\left(\frac{1}{2}-\frac{\cos 2 \theta}{2}\right) d \theta$

$=\frac{1}{2 \times 2} \int_{0}^{\frac{\pi}{2}} 1 d \theta-\frac{1}{4} \int_{0}^{\frac{\pi}{2}} \cos 2 \theta d \theta$

$\left[\because \int 1 d x=x\right]$

$\left[\because \int \cos a x d x=\frac{\sin a x}{a}\right]$

$=\frac{1}{4}[\theta]_{0}^{\frac{\pi}{2}}-\frac{1}{4}\left[\frac{\sin 2 \theta}{2}\right]_{0}^{\frac{\pi}{2}}$

$=\frac{1}{4}\left[\frac{\pi}{2}-0\right]-\frac{1}{8}[\sin 2 \theta]_{0}^{\frac{\pi}{2}}$

$=\frac{1}{4} \frac{\pi}{2}-\frac{1}{8}\left[\sin 2 \times \frac{\pi}{2}-\sin 2 \times 0\right]$

$=\frac{\pi}{8}-\frac{1}{8}[\sin \pi-\sin 0]$

$=\frac{\pi}{8}-\frac{1}{8}[0-0]\; \; \; \; \; \; \; \; \; \; \quad[\because \sin \pi=\sin 0=0]$

$=\frac{\pi }{8}$