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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 42 Maths Textbook Solution.

Answers (1)

Answer:

$\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C^{\prime}$

$\text { where } C^{\prime}=C-\ln a$

Given:

$\int \frac{1}{\sqrt{x^{2}-a^{2}}} d x$

Hint:

In this statement we have to assume x as asecθ.

Solution:

$I=\int \frac{1}{\sqrt{x^{2}-a^{2}}} d x$

$\text { putting } x=a \sec \theta$

$\Rightarrow d x=a \sec \theta \tan \theta d \theta$

$I=\int \frac{a \sec \theta \tan \theta d \theta}{\sqrt{a^{2} \sec ^{2} \theta-a^{2}}}$

$=\int \frac{a \sec \theta \tan \theta d \theta}{a \cdot \tan \theta}$

$=\int \sec \theta \tan \theta d \theta$

$=\ln |\sec \theta+\tan \theta|+C$

$=\ln \left|\sec \theta+\sqrt{\sec ^{2} \theta-1}\right|+C$

$=\ln \left|\frac{x}{a}+\sqrt{\left(\frac{x}{a}\right)^{2}}-1\right|+C$

$=\ln \left|\frac{x+\sqrt{x^{2}-a^{2}}}{a}\right|+C$

$=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C$

$=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|-\ln a+C$

$=\ln \left|x+\sqrt{x^{2}-a^{2}}\right|+C^{\prime}$

$\text { where } C^{\prime}=C-\ln a$

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