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Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 116 Maths Textbook Solution.

Answers (1)

Answer:

2\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]+C

Hint: to solve this statement we will suppose x as \sin \theta

Given: \int \cos ^{-1}(1-2 x)^{2} d x

Solution:

\int \cos ^{-1}(1-2 x)^{2} d x

\operatorname{let} x=\sin \theta, d x=\cos \theta d \theta

I=\int \cos ^{-1}\left(1-2 \sin ^{2} \theta\right)^{2} \cos \theta d \theta

I=\int \cos ^{-1}(\cos 2 \theta) \cos \theta d \theta

I=\int 2 \theta \cos \theta d \theta

   \int u v d x=u \int v d x-\int\left[\frac{d u}{d v} \int v d x\right] d x

  I=2\left[\theta \sin \theta-\int \sin \theta d \theta\right]

I=2[\theta \sin \theta+\cos \theta]+c

I=2\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]+C

 

 

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