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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 112 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 112 Maths Textbook Solution.

Answers (1)

Answer:\frac{1}{2}\left[x \cos ^{-1} x-\sqrt{1-x^{2}}\right]+C

Hint: to solve this question we will presume x as \cos \theta

Given: \int \tan ^{-1} \sqrt{\frac{1-x}{1+x}} d x

Solution:

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

\operatorname{let} x=\cos 2 \theta

\frac{d x}{d \theta}=-2 \sin 2 \theta

d x=-2 \sin 2 \theta d \theta

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

\cos 2 \theta=2 \cos ^{2} \theta-1

\cos 2 \theta=1-2 \sin ^{2} \theta

I=\int \tan ^{-1} \sqrt{\frac{1-1+2 \sin ^{2} \theta}{1+2 \cos ^{2} \theta-1}} \times-2 \sin 2 \theta d \theta

I=\int \tan ^{-1} \sqrt{\tan ^{2} \theta} \times-2 \sin 2 \theta d \theta

I=-2 \int \theta \sin 2 \theta d \theta

I=-2\left[\theta \int \sin 2 \theta d \theta-\int \frac{d \theta}{d \theta} \int \sin 2 \theta d \theta\right] d x . \text { .applying Byparts }

I=-2\left[\theta \times \frac{-\cos 2 \theta}{2}-(-1) \int \frac{\cos 2 \theta}{2}\right] d \theta

I=-2\left[-\theta \frac{\cos 2 \theta}{2}-(-1) \int \frac{\cos 2 \theta}{2}\right] d \theta

I=-2\left[-\theta \frac{\cos 20}{2}+\int \frac{\cos 20}{2}\right] d \theta

I=-2\left[-\frac{\cos 2 \theta}{2}+\frac{\sin 2 \theta}{4}\right]+C

I=-2\left[\left(\frac{-1}{2} \cos ^{-1} x\right) \frac{x}{2}+\frac{\sin \left(\cos ^{-1} x\right)}{4}+C\right]

I=-2\left[\frac{-x}{4} \cos ^{-1} x+\frac{\sin \left(\sin ^{-1} \sqrt{1-x^{2}}\right)}{4}\right]+C

I=-\frac{2}{4}\left[-x \cos ^{-1} x+\sqrt{1-x^{2}}\right]+C

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

 

 

Posted by

infoexpert21

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