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

Question

Got the following integral, have no clue:

$$ \int_{}^{} x^2\sqrt{a^2+x^2} \; dx $$

What I tried to do:

1. $$t=a^2+x^2$$

$$dt = 2x\;dx$$

$$x=\sqrt{t-a^2} $$ 2. Got the following integral and got stuck: $$ \frac{1}{2}\int_{}^{} \sqrt{t}\sqrt{t-a^2} \; dt $$

Answer 1

Hint:

Try to evaluate the integral by substituting $x=a\tan\theta$, so that \begin{align} dx &= a \sec^{2} \theta \\ a^{2}+x^{2} &= a^{2}(1+\tan^{2} \theta)\\ &= a^{2} \sec^{2} \theta \\ & \vdots \end{align}

Alternatively, alluding to @Did's comment below \begin{align} x &= \ \sinh \theta \\ \implies dx &= a \cosh \theta d \theta\\ \sqrt{a^{2}+x^{2}} &= a\sqrt{1+\sinh^{2} \theta} \\ &= a \cosh \theta \\ \implies \int x^{2} \sqrt{a^{2}+x^{2}}dx &= \int a^{4}\sinh^{2}\theta \cosh^{2} \theta d \theta\\ &\vdots \end{align}

Answer 2

$x=a\sinh{t}$ also works: $dx=a\cosh{t}dt$...