Integration of the rational function $ 4/( 1+4t^2) $ So it's August so my memory of math is a little rough right now. I was wondering if someone could help me with integration with a fraction involved?
For example:
$$\int_0^{1/2} \frac 4{1+4t^2} \, dt$$
Note: Yeah, this is a homework question, and I already know the answer is $\frac 12 \pi$, but I would really appreciate it if someone could help me through this problem. Could it be $u$-substitution?
 A: (Note: the perspective offered in this response is surely more than what is expected from the OP's homework problem. This answer is instead for those interested in a more rigorous approach.)
As was correctly indicated in the question statement, the following integral is indeed equal to $\frac12\pi$:
$$I=\int_{0}^{1/2}\frac{4\,\mathrm{d}t}{1+4t^2}=\frac12\pi.$$
We'd like to rigorously prove that $I=\frac12\pi$, and in order to do so we should first adopt a rigorous definition of $\pi$ (preferably defined as an integral). I prefer to define $\pi$ by the integral,
$$\pi:=\int_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{1-x^2}},$$
because this integral represents the circumference-to-diameter ratio of a unit circle.

Rescaling the integral by the substitution $u=2t$, the integral becomes:
$$\begin{align}
I
&=\int_{0}^{1/2}\frac{4\,\mathrm{d}t}{1+4t^2}\\
&=2\int_{0}^{1/2}\frac{2\,\mathrm{d}t}{1+(2t)^2}\\
&=2\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}.
\end{align}$$
Using the reciprocal substitution $w=\frac{1}{u}$, we see that the integral from $0$ to $1$ of $\frac{1}{1+u^2}$ also equals the integral from $1$ to $\infty$ of $\frac{1}{1+u^2}$:
$$\begin{align}
\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}
&=\int_{\infty}^{1}\frac{\frac{(-1)}{w^2}\mathrm{d}w}{1+\left(\frac{1}{w}\right)^2}\\
&=\int_{1}^{\infty}\frac{\frac{1}{w^2}\mathrm{d}w}{1+\frac{1}{w^2}}\\
&=\int_{1}^{\infty}\frac{\mathrm{d}w}{w^2+1}\\
&=\int_{1}^{\infty}\frac{\mathrm{d}u}{1+u^2}.
\end{align}$$
Hence,
$$\begin{align}
I
&=2\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}\\
&=\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}+\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}
\\
&=\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}+\int_{1}^{\infty}\frac{\mathrm{d}u}{1+u^2}\\
&=\int_{0}^{\infty}\frac{\mathrm{d}u}{1+u^2}\\
&=\frac12\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{1+u^2}.
\end{align}$$
Finally, using the substitution $u=\frac{x}{\sqrt{1-x^2}}$, the integral becomes:
$$\begin{align}
I
&=\frac12\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{1+u^2}\\
&=\frac12\int_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{1-x^2}}\\
&=\frac12\pi.
\end{align}$$
A: $$\int_{0}^{\frac{1}{2}}\frac{4}{1+4t^{2}}dt=4\int_{0}^{\frac{1}{2}}\frac{1}{1+(2t)^{2}}dt=2\int_{0}^{1}\frac{1}{1+u^{2}}du$$
where I have used the substitution $u=2t$ so that $du=2dt$. Now let $u=\tan(\theta)$ so that $du=\sec^{2}(\theta)d\theta$ then we get:
$$=2\int_{0}^{\frac{\pi}{4}}\frac{\sec^{2}(\theta)}{1+\tan^{2}(\theta)}d\theta=2\int_{0}^{\frac{\pi}{4}}d\theta=2\frac{\pi}{4}=\frac{\pi}{2}$$
