Calculus (Integration) Is there a simple way to integrate $\displaystyle\int\limits_{0}^{1/2}\dfrac{4}{1+4t^2}\,dt$
I have no idea how to go about doing this. The fraction in the denominator is what's confusing me. I tried U-Substitution to no avail.
 A: $$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt$$
We set $t=\frac{\tan{u}}{2}$ we have the following:
$t=0: u=0$
$t=\frac{1}{2}: u=\frac{\pi}{4}$
$dt=\frac{1}{2 \cos^2{u}}du$
$$\frac{4}{1+4t^2}=\frac{4}{1+4 \frac{\tan^2{u}}{4}}=\frac{4}{1+\tan^2{u}}=\frac{4 \cos^2{u}}{\sin^2{u}+\cos^2{u}}=4 \cos^2{u}$$
Therefore, we have the following:
$$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt=\int_0^{\frac{\pi}{4}}2 du=2\frac{\pi}{4}=\frac{\pi}{2}$$
A: $$
\int_0^{\frac{1}{2}} \frac{4}{4t^2+1}\ dt
$$
Let $u=2t$, 
$$ \frac{d}{dt}u=\frac{d}{dt}[2t]=2 \Rightarrow  du = 2\ dt $$
$$
\int_0^1 \frac{2}{u^2+1}\ du= 2 \int_0^1 \frac{1}{u^2+1}\ du=2\arctan u\bigg|_0^1
$$
$$
=2\arctan 1 -2\arctan 0=2\frac{\pi}{4}-0=\frac{\pi}{2}
$$
A: A slightly different way to find this integral is as follows:
$\displaystyle\int_{0}^{\frac{1}{2}}\frac{4}{4t^2+1}\; dt=\frac{4}{4}\int_{0}^{\frac{1}{2}}\frac{1}{t^2+\frac{1}{4}}\;dt=\frac{1}{\frac{1}{2}}\left[\arctan\frac{t}{\frac{1}{2}}\right]_{0}^{\frac{1}{2}}=2\left[\arctan 2t\right]_{0}^{\frac{1}{2}}$
$\displaystyle=2(\arctan1-\arctan0)=2(\frac{\pi}{4}-0)=\frac{\pi}{2}$, $\;\;$using the formula $\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$.
