Indefinite integral - What am I doing wrong? We have the integral $\displaystyle \int \sqrt{4-x^2} dx$, and I want to evaluate it using integration by parts. So our first step gives us:
$$ \int \sqrt{4-x^2} dx = x \sqrt{4-x^2} + \int \dfrac{x^2}{\sqrt{4-x^2}} dx   $$
Now we have to deal with $\displaystyle \int \dfrac{x^2}{\sqrt{4-x^2}} dx $. I do integration by parts again: 
$$\displaystyle \int \dfrac{x^2}{\sqrt{4-x^2}} dx = x^2 \displaystyle \int \dfrac{1}{\sqrt{4-x^2}} dx -2 \displaystyle \int \dfrac{x}{\sqrt{4-x^2}} dx $$
We have $\displaystyle \int \dfrac{1}{\sqrt{4-x^2}} dx = \arcsin(\dfrac{x}{2})$ and $\displaystyle \int \dfrac{x}{\sqrt{4-x^2}} dx = -\sqrt{4-x^2}$, we have $$\displaystyle \int \sqrt{4-x^2} dx = x \sqrt{4-x^2} +x^2 \arcsin(\dfrac{x}{2}) + 2\sqrt{4-x^2} +C $$ 
However, this is incorrect. Can anyone tell me where I made a mistake?

Usually for this form of integrand ($\sqrt{a^2-x^2}$) we use a trig sub with $x=a\sin\theta$. In our case $a=2$. 
$$x=2\sin\theta, dx=2\cos\theta d\theta$$
Substituting we have
$$\int\sqrt{4-x^2}dx=\int2\cos\theta\sqrt{4-4\sin^2\theta}\,d\theta$$
$$=4\int\cos^2\theta\, d\theta$$
We can rewrite $\cos^2\theta$ as $\cos(2\theta)/2+1/2$. 
$$=2\int\cos2\theta\,d\theta+2\theta$$
Let $u=2\theta$, $du=d\theta$, then
$$=\int\cos u \,du+u$$
$$=u+\sin u+C$$
$$=2\theta+\sin2\theta+C$$
Now we use the double angle formula.
$$=2\theta+2\sin\theta\cos\theta+C$$
$$=2\theta+2\sin\theta\sqrt{1-\sin^2\theta}+C$$
Solving for theta in our previous substitution we have $\theta=\arcsin x/2$. Therefore
$$\int \sqrt {4-x^2}=2\arcsin \left(\frac x 2\right) +\frac x 2 \sqrt{4-x^2}+C$$
