How to compute $\int\frac{x^2}{x^2+4}\,dx$ and $\int\frac{1}{\sqrt{4-x^2}}\,dx$? My calculus teacher gave us a list of integrals to solve through the substitution method. I've been trying everything for hours, and I just can't find out how to compute these:
$$∫\frac{x^2}{x^2+4}\,dx\qquad\text{and}\qquad \int\frac{1}{\sqrt{4-x^2}}\,dx.$$
Is it even possible to solve them through substitution?
 A: The first one can be solved by rewriting it first
$$\int\frac{x^2}{x^2+4}\,dx = \int\frac{x^2+4-4}{x^2+4}\,dx = \int\,dx - 4\int\frac{dx}{x^2+4}.$$
Then we can solve the second one of these integrals (first one is easy) by factoring out $4$ from the denominator and then using the substitution $u=\frac{x}{2}$:
$$\begin{align*}
\int\frac{dx}{x^2+4} & = \frac{1}{4}\int\frac{dx}{\left(\frac{x}{2}\right)^2+1}\\
&= \frac{1}{2}\int\frac{du}{u^2+1}.\end{align*}$$
The last integral has a direct antiderivative.
For the second integral, try $x=2u$. Then $dx = 2du$, so
$$\int\frac{1}{\sqrt{4-x^2}}\,dx = \int\frac{2du}{\sqrt{4-4u^2}} = \int\frac{du}{\sqrt{1-u^2}}.$$
Does the last one look familiar?
Alternatively, do the same trick: factor out $4$ in the denominator inside the square root:
$$\begin{align*}
\int\frac{1}{\sqrt{4-x^2}}\,dx &= \int\frac{1}{\sqrt{4(1 - (\frac{x}{2})^2}}\,dx\\
&= \int\frac{1}{2\sqrt{1 - (\frac{x}{2})^2}}\end{align*}$$
which may suggest the substitution $w=\frac{x}{2}$.

Added. In light of the comments...
My first impulse with the second integral would have been to use a trigonometric substitution. That's because integrals that involve $\sqrt{a^2-x^2}$ but which do not have an $x\,dx$ factor are the traditional proving grounds for the trigonometric substitution. Here, $a=2$, so we would try the substitution $x=2\sin\theta$, with $0\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, $dx = 2\cos\theta\,d\theta$. This wold turn the $\sqrt{4-x^2}$ factor into
$$\sqrt{4-x^2} = \sqrt{4-4\sin^2\theta} = \sqrt{4(1-\sin^2\theta)} = \sqrt{4\cos^2\theta} = 2|\cos\theta|=2\cos\theta$$
with the last equality because $-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}$, so $\cos\theta\geq 0$. Then the integral would simplify to
$$\int\frac{dx}{\sqrt{4-x^2}} = \int\frac{\cos\theta\,d\theta}{2\cos\theta} = \frac{1}{2}\int\,d\theta,$$
which is trivial to do; finally, we would return to $x$ using $\theta=\frac{1}{2}\arcsin(x)$. 
A: For the second one, use this substitution if you like:
$x=2 \sin u$. Then $dx=2\cos u\ du$.Substituting that into the second integral,we can reduce the problem like this :
$$\int \frac{1}{\sqrt{4-x^2}}dx=\int du=u+C$$[where C is the constant of integration]$$=\sin^{-1}\frac{x}{2}+C$$
