Assuming that 0≤θ<π/2, here is the integration that I am required to evaluate:
I have been successful in solving the majority of this integral, however, the final answer is as follows:
I am unsure on how to obtain this answer, and I have been guided by my teacher that it involved the use of right-angled triangles. Regardless if you have a method involving right-angled triangles or not, it would be highly appreciated if anyone could please help me in reaching the final answer.
Please note: I have tried using MathJax but unfortunately have struggled with this programme, so I would be grateful if you could please excuse my use of images in this question :)
Consider $f(x)=\frac{x^2}{\sqrt{4-x^2}}$. It is suggested that you set $x=2\sin(u)$ as a change of variable. $$f(x)=\frac{x^2}{\sqrt{4-x^2}}=\frac{4\sin^2(u)}{\sqrt{4-4\sin^2(u)}}=\frac{4\sin^2(u)}{\sqrt{4(1-\sin^2(u))}}$$ We know that $\sin^2(u)+\cos^2(u)=1$, thus $1-\sin^2(u)=\cos^2(u)$ So, $$\frac{4\sin^2(u)}{\sqrt{4\cos^2(u)}}=\frac{4\sin^2(u)}{\sqrt{(2\cos(u))^2}}=\frac{2\sin^2(u)}{\cos(u)}$$
We know proceed to make the change of variable:
$x=2sin(u)\implies dx=2\cos(u).du\ $, and $$\int f(x)\ dx=2\int\frac{\sin^2(u)}{cos(u)}2cos(u)\ du=4\int \sin^2(u)\ du$$ A common wa to deal with $\int \sin^2(u)\ du$ is to use the trigonometric relation $\sin^2(u)=\frac{1}{2}-\frac{1}{2}\cos(2u)$ as this is easier to integrate: $$4\int\sin^2(u)\ du=4\int\frac{1}{2}-\frac{1}{2}\cos(2u)\ du = *$$ Now we simplify and solve : $$*=2\int du-2\int\cos(2u)\ du=2u-\sin(2u)+C=2u-2\sin(u)\cos(u)+C$$ By applying the trigonometric relation: $\sin(2u)=2\sin(u)\cos(u)$. And finally, by reverting $\cos$ into $\sin$ using $\cos^2(u)=1-\sin^2(u)$: $$2u-2\sin(u)\cos(u)+C=2u-2\sin(u)\sqrt{1-\sin^2(u)}+C$$
Now, we substitute back $x=2\sin(u)\implies u=\sin^{-1}(\frac{x}{2})$ and $$\int\frac{x^2}{\sqrt{4-x^2}}\ dx=2u-2\sin(u)\sqrt{1-\sin^2(u)}+C=2\sin^{-1}(\frac{x}{2})-\frac{1}{2}x\sqrt{4-x^2}+C$$ Using a few more manipulation to transform $\sin^2(u)$
Hope this helps.
Edit: Quick manipulation:
$$2\sin(u)\sqrt{1-\sin^2(u)}=2\sin\big(\sin^{-1}(\frac{x}{2})\big)\sqrt{1-\sin\big(\sin^{-1}(\frac{x}{2})\big)^2}$$ $$=x\sqrt{1-\frac{x^2}{4}}=x\sqrt{\frac{4-x^2}{4}}=\frac{x}{2}\sqrt{4-x^2}$$