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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 3

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Answer: -\frac{1}{3}\cos x^{3}+c

Hint: You must know about the integration value sin function

Given:  \int x^{2}\sin x^{3}dx

Explanation:  \int x^{2}\sin x^{3}dx

Let x^{3}=t  and differentiate both sides,


=\frac{1}{3}\int \sin t dt

=-\frac{1}{3} \cos t +c

=-\frac{1}{3} \cos x^{3} +c                                       \left [ \int \sin x dx= -\cos x+c \right ]

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