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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 60 Maths Textbook Solution.

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Answer: \frac{1}{2}\left [ -\sin ^{-1}\left ( x \right )^{2}\sqrt{\left ( 1-x^{4} \right )+x^{2}} \right ]+c

Hint: Let \sin ^{-1}\left ( x^{2} \right )=t

Given: \int \frac{x^{3}\sin ^{-1}x^{2}}{\sqrt{1-x^{4}}}dx

Solution: Let \sin ^{-1}\left ( x^{2} \right )=t\Rightarrow \frac{2x}{\sqrt{1-x^{4}}}dx=dt

                \int \frac{x^{3}\sin ^{-1}x^{2}}{\sqrt{1-x^{4}}}dx

                \begin{aligned} &=\frac{1}{2} \int \frac{2 x \sin ^{-1}\left(x^{2}\right) \times x^{2}}{\sqrt{1-x^{4}}} d x \\ &=\frac{1}{2} \int t \sin t d t \\ &=\int\left[t \int \sin t d t+\int \cos t d t\right]+c \\ &=\frac{1}{2}[-t \cos t+\sin t]+c \\ &=\frac{1}{2}\left[-\sin ^{-1}(x)^{2} \sqrt{\left(1-x^{4}\right)}+x^{2}\right]+c \end{aligned}

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