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?

  • $\begingroup$ If you are interested in how this particular formula is typeset, see here. For general advice, you can consult the notation help page $\endgroup$
    – AlexR
    Commented Aug 7, 2014 at 3:25
  • 1
    $\begingroup$ This looks a lot like $\frac{1}{1+x^2}$, and we know $\int \frac{1}{1+x^2}=\arctan x+C$. So let $2t=x$. $\endgroup$ Commented Aug 7, 2014 at 3:27

2 Answers 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 $\frac12\pi$:


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,


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 $\frac{1}{1+u^2}$:

$$\begin{align} \int_{0}^{1}\frac{\mathrm{d}u}{1+u^2} &=\int_{\infty}^{1}\frac{\frac{(-1)}{w^2}\mathrm{d}w}{1+\left(\frac{1}{w}\right)^2}\\ &=\int_{1}^{\infty}\frac{\frac{1}{w^2}\mathrm{d}w}{1+\frac{1}{w^2}}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}w}{w^2+1}\\ &=\int_{1}^{\infty}\frac{\mathrm{d}u}{1+u^2}. \end{align}$$


$$\begin{align} I &=2\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}\\ &=\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}+\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2} \\ &=\int_{0}^{1}\frac{\mathrm{d}u}{1+u^2}+\int_{1}^{\infty}\frac{\mathrm{d}u}{1+u^2}\\ &=\int_{0}^{\infty}\frac{\mathrm{d}u}{1+u^2}\\ &=\frac12\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{1+u^2}. \end{align}$$

Finally, using the substitution $u=\frac{x}{\sqrt{1-x^2}}$, the integral becomes:

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



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:


  • $\begingroup$ How did you think to do that? In my pre-calc honors class we only did questions with u-sub where the u would be something visible in the problem without having to pull anything out. $\endgroup$
    – Hello
    Commented Aug 7, 2014 at 3:32
  • $\begingroup$ When you let $u=f(x)$ notice that this can be rephrased as $x=f^{-1}(u)$ provided the function is invertible (which it has to be in order to apply a change of variables). So the second substitution is really letting $\theta=\arctan(u)$ and you can rewrite the expression as: $\frac{1}{1+u^{2}}=\frac{1}{1+(\tan(\arctan(u)))^{2}}$ so that it is visible. $\endgroup$
    – user71352
    Commented Aug 7, 2014 at 3:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .