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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 86 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 86 Maths Textbook Solution.

Answers (1)

Answer:

a^{2}\left(\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\left(\frac{x}{a}\right) \cdot \sqrt{a^{2}-x^{2}}\right)+c

Hint:

You must know about how to solve integration.

Given:

\int \sqrt{a^{2}-x^{2}} d x

Solution:

\text { let } x=a \sin \theta

   d x=a \cos \theta d \theta

I=\int \sqrt{a^{2}-a^{2} \sin ^{2} \theta} d \theta

I=a^{2} \int \sqrt{1-\sin ^{2} \theta} \cdot \cos \theta d \theta \quad \sqrt{1-\sin ^{2} \theta}=\cos ^{2} \theta

I=a^{2}\left(\frac{\theta}{2}+\frac{1}{2} \frac{(\sin 2 \theta)}{2}\right)+c                                        \left[\operatorname{let} \theta=\sin ^{-1}\left(\frac{x}{a}\right), \cos \theta=\sqrt{1-\left(\frac{x^{2}}{a^{2}}\right)}\right.

I=a^{2}\left(\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\frac{1}{2} \frac{2(\sin 2 \theta) \cos \theta}{2}\right)+c

I=a^{2}\left(\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\frac{1}{2}\left(\frac{x}{a}\right) \cdot \frac{1}{a} \sqrt{a^{2}-x^{2}}\right)+c

I=a^{2}\left(\frac{1}{2} \sin ^{-1}\left(\frac{x}{a}\right)+\left(\frac{x}{a}\right) \cdot \sqrt{a^{2}-x^{2}}\right)+c

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