Mathematics

# Prove that $\displaystyle \int \sqrt{x^2 - a^2} dx = \dfrac{x}{2} \sqrt{x^2 - a^2} - \dfrac{a^2}{2} \log |x + \sqrt{x^2 - a^2}| + c$.

##### SOLUTION
Given, $\displaystyle \int \sqrt{x^2 - a^2} dx$
$= \sqrt{x^2 - a^2}(x) -\displaystyle \int \dfrac{2x}{2 \sqrt{x^2 - a^2}} x dx$
$= \sqrt{x^2 - a^2}(x) - \displaystyle \int \dfrac{x^2}{\sqrt{x^2- a^2}} dx$
$= \sqrt{x^2 - a^2}(x) - \displaystyle \int \dfrac{x^2 - a^2 + a^2}{\sqrt{x^2 - a^2}} dx$
$= \sqrt{x^2 - a^2} (x) -\displaystyle \int \dfrac{x^2 - a^2}{\sqrt{x^2 - a^2}} dx + a^2 \int \dfrac{1}{\sqrt{x^2 - a^2}} dx$
$= \sqrt{x^2 - a^2} (x) - \displaystyle \int \sqrt{x^2 - a^2} dx + a^2 \int \dfrac{1}{\sqrt{x^2 - a^2}} dx$
$2 \displaystyle \int \sqrt{x^2 - a^2} dx = \sqrt{x^2 - a^2} (x) + a^2 \int \dfrac{1}{\sqrt{x^2 - a^2}} dx$
$\displaystyle \int \sqrt{x^2 - a^2} dx = \dfrac{1}{2} \sqrt{x^2 - a^2}(x) + \dfrac{1}{2} a^2 \log |x + \sqrt{x^2 - a^2}| + c$   ....where $c$ is constant term.

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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