Integrating $\sqrt{x^{2}+a^{2}}$ and $\sqrt{x^{2}-a^{2}}$

Question

How can I integrate the square rooted function? I’m having trouble regarding the substitution by a trigonometric ratio…

$$\sqrt{a^{2}+x^{2}}$$

And

$$\sqrt{x^{2}-a^{2}}$$

Answer 1

$$a^{2}+x^{2}$$ $$*Substituting x=atan\theta , we get:*$$ $$a^{2}+a^{2}tan^{2}\theta$$ $$=a^{2}(1+tan^{2}\theta)$$ $$=a^{2}sec^{2}\theta$$

In the square root , $$=asec\theta$$

Again , $$dx=asec^{2}\theta d\theta...$$ So , the integral becomes :

$$∫a^{2}sec^{3}\theta$$ $$=∫a^{2}sec\theta(1+tan^{2}\theta)$$ $$=a^{2}∫sec\theta+a^{2}∫sec\theta.tan^{2}\theta$$ $$∫sec\theta=ln(sec\theta+tan\theta)$$ $$sec\theta.tan^{2}\theta=$$ $$=tan\theta.dsec\theta$$

Then apply substitution by parts...

The second integral can be solved by looking at the last part of the first one...

Answer 2

Try using $x=a \sinh(u)$ as $a^2 + a^2 \sinh^2(u) = a^2 \cosh^2(u)$.

Try $x= a \sec(u)$ for the other. In each case you are exploiting a trig identity which you know (such as $\tan^2(u) + 1 = \sec^2(u)$) and multiplying through by $a$.