Integration of the rational function $ 4/( 1+4t^2) $

Question

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?

Answer 1

$$\int_{0}^{\frac{1}{2}}\frac{4}{1+4t^{2}}dt=4\int_{0}^{\frac{1}{2}}\frac{1}{1+(2t)^{2}}dt=2\int_{0}^{1}\frac{1}{1+u^{2}}du$$

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:

$$=2\int_{0}^{\frac{\pi}{4}}\frac{\sec^{2}(\theta)}{1+\tan^{2}(\theta)}d\theta=2\int_{0}^{\frac{\pi}{4}}d\theta=2\frac{\pi}{4}=\frac{\pi}{2}$$

Answer 2

(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.

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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:

$$\begin{align} I &=\frac12\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{1+u^2}\\ &=\frac12\int_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{1-x^2}}\\ &=\frac12\pi. \end{align}$$