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
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