<s>[INST] 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.
 [/INST] $$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt$$
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:
$$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt=\int_0^{\frac{\pi}{4}}2 du=2\frac{\pi}{4}=\frac{\pi}{2}$$
</s>