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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 35 Maths Textbook Solution.

Answers (1)

Answer: 3\sin ^{-1}x,x+3\sqrt{1-x^{2}}+c

Given: \int \sin ^{-1}\left ( 3x-4x^{3} \right )dx

Hint:

        3x-4x^{3}\rightarrow \sin \theta

         x=\sin \: t

Solution:

            I=\int \sin ^{-1}\left ( 3x-4x^{3} \right )dx

                                Let x = sint  =>dx = costdt & 3\: \sin t-4\sin ^{3}t=\sin 3t

           \begin{aligned} &=\int \sin ^{-1} \sin 3 t \cos t d t \\ &=\int 3 t \cos t d t \\ &=3 \int t \cos t d t \\ &I=3\left[\left(t \int \cos t d t-\int(1 \sin t) d t\right)\right] \\ &=3\left[t \sin t-\int \sin t d t\right] \\ &=3[t \sin t-(-\cos t+c)] \\ &=3 t \sin t+3 \cos t+c \\ &=3 \sin ^{-1} x \cdot x+3 \sqrt{1-x^{2}}+c \end{aligned}

 

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