How to evaluate the definite integral $\int_0^{a/2 }x^2(a^2-x^2)^{-3/2}dx $

Question

$$\int_0^{a/2}x^2(a^2-x^2)^{-3/2}dx $$

I've been having trouble with this question for a while now. When I attempted the question I substituted $x=a\sin\theta$ and $(a\cos\theta)^3=(a^2-x^2)^{-3/2}$. When I solved with the substitutions what I get at the end is $\tan^2\theta$ in the integral sign. So I integrated $\sec^2\theta-1$ and got an answer which is different to the real answer. And when I checked an online integral calculator the top $x^2$ was split into $x^2-a^2+a^2$ to make two fractions. which didn't make sense to me at all because somewhere along that line the $x^2$ itself was completely removed. The online calculator is www.integral-calculator.com I learned trigonometric substitution last night and I am not very good at it. So a detailed explanation would be very much appreciated.

Answer 1

Hint: Using substitution $x=a\sin t$ then $$\int_0^{\frac{a}{2}}\dfrac{x^2}{\sqrt{a^2-x^2}^3}dx=\int_0^{\pi/6}\tan^2t\ dt$$

Answer 2

Hint:

How about integration by parts

$$\int x\cdot\dfrac x{(a^2-x^2)^{3/2}}dx=x\int\dfrac x{(a^2-x^2)^{3/2}}dx-\int\left(\dfrac{dx}{dx}\int\dfrac x{(a^2-x^2)^{3/2}}dx\right)dx$$

Now choose $\sqrt{a^2-x^2}=y$