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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 115 Maths Textbook Solution.

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Answer:\left(\sin ^{-1} x\right)^{3} \cdot x+3\left(\sin ^{-1} x\right)^{2} \sqrt{1-x^{2}}+6 \sin ^{-1} x-6 \sqrt{1-x^{2}}+C

Hint: to solve this equation, we have to convert equation into trigonometric formulaes

Given: \int\left(\sin ^{-1} x\right)^{3} d x


\text { Let } \sin ^{-1} x=\theta

x=\sin \theta=>d x=\cos \theta d \theta

I=\int \theta^{3} \cdot \cos \theta d \theta

I=\theta^{3} \sin \theta-3 \int \theta^{2} \sin \theta d \theta

I=\theta^{3} \sin \theta+3 \theta^{2} \cos \theta-\left(60(-\sin \theta)-\int 6(-\sin \theta)\right.

I=\theta^{3} \sin \theta+3 \theta^{2} \cos \theta+60 \sin \theta-6 \cos \theta+C

I=\left(\sin ^{-1} x\right)^{3} \cdot x+3\left(\sin ^{-1} x\right)^{2} \sqrt{1-x^{2}}+6 \sin ^{-1} x-6 \sqrt{1-x^{2}}+C

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