#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 56 Maths Textbook Solution.

Answer: $\frac{2}{3}$

Hint: Use indefinite formula then put the limit to solve this integral

Given:$\int_{0}^{\frac{\pi }{2}}\sin ^{3}xdx$

Solution: Let,$I=\int_{0}^{\frac{\pi }{2}}\sin ^{3}xdx=\int_{0}^{\frac{\pi }{2}}\sin ^{2}x\sin xdx$

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

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

$=\int_{0}^{\frac{\pi}{2}} \sin x d x-\int_{0}^{\frac{\pi}{2}} \cos ^{2} x \sin x d x$

Put$t=\cos x \Rightarrow d t=-\sin x d x \Rightarrow d x=\frac{-d t}{\sin x}$ in the 2nd Integral term

$=\int_{0}^{\frac{\pi}{2}} \sin x d x-\int_{0}^{\frac{\pi}{2}} t^{2} \sin x \frac{d t}{-\sin x}$

$=\int_{0}^{\frac{\pi}{2}} \sin x d x+\int_{0}^{\frac{\pi}{2}} t^{2} d t$                                                                    $\quad\left[\int \sin x d x=-\cos x, \int x^{n} d x=\frac{x^{n+1}}{n+1}\right]$

$=[-\cos x]_{0}^{\frac{\pi}{2}}+\left[\frac{t^{2+1}}{2+1}\right]_{0}^{\frac{\pi}{2}}$

$=-\left[\cos \frac{\pi}{2}-\cos 0\right]+\frac{1}{3}\left[t^{3}\right]_{0}^{\frac{\pi}{2}}$

$=-\left[\cos \frac{\pi}{2}-\cos 0\right]+\frac{1}{3}\left[\cos ^{3} x\right]_{0}^{\frac{\pi}{2}}$                                                        $\left [ \cos \frac{\pi }{2}=0,\cos 0=1 \right ]$

$=-[0-1]+\frac{1}{3}\left[\cos ^{3} \frac{\pi}{2}-\cos ^{3} 0\right]_{0}^{\frac{\pi}{2}}$

$=1+\frac{1}{3}\left[0^{3}-1^{3}\right]=1+\frac{1}{3}[0-1]$

$=1-\frac{1}{3}=\frac{3-1}{3}=\frac{2}{3}$