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

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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

Putt=\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}}



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