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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 54 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 54 Maths Textbook Solution.

 

Answers (1)

Answer:

I=\sin ^{-1} \sqrt{x}+\frac{1}{2} \sin 2 \sin ^{-1} \sqrt{x}+c

Given:

\int \sqrt{\frac{1-x}{x}} d x

Hint:

To solve this question we have to assume that x as sin²θ

Solution: 

I=\int \sqrt{\frac{1-x}{x}} d x

x=\sin ^{2} \theta \quad d x=2 \sin \theta \cos \theta d \theta

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

I=\int \frac{\sqrt{\cos ^{2} \theta}}{\sqrt{\sin ^{2} \theta}}(2 \sin \theta \cos \theta) d \theta

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

I=\int 2 \cos ^{2} \theta d \theta

I=\int(1+\cos 2 \theta) d \theta

I=\int 1 d \theta+\int \cos 2 \theta d \theta

I=\theta+\frac{1}{2} \sin 2 \theta+c

I=\sin ^{-1} \sqrt{x}+\frac{1}{2} \sin 2 \sin ^{-1} \sqrt{x}+c

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