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

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

Answers (1)

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}}

=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}

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