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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 41 Maths Textbook Solution.

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

Hint: \cos 3 \: x=4\cos ^{3}x-3\: \cos \: x

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

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

             Let x=\cos \theta \Rightarrow \theta =\cos ^{-1}x

             \begin{aligned} &I=\int \cos ^{-1}\left(4 \cos ^{3} \theta-3 \cos \theta\right) d x \\ &=\int \cos ^{-1}(\cos 3 \theta) d x \\ &=\int 3 \theta d x \\ &=3 \int \cos ^{-1} x d x \end{aligned}

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


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