# 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.

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Note that $\displaystyle \int \dfrac4{1+4t^2}\mathrm dt=2\int \dfrac 2{1+(2t)^2}\mathrm dt$ and recall that $\displaystyle \int \dfrac{u'}{1+u^2}=\ldots$. –  Git Gud Aug 27 '14 at 23:06
You can factor out the 4 in the bottom and use the arctangent formula, or substitute u=2t and then use this formula. –  user84413 Aug 27 '14 at 23:08
I haven't learned that formula. I just finished precalc and my AP calc summer assignment is an AP exam so I haven't learned everything. The teacher said that might happen. thanks –  Hello Aug 27 '14 at 23:09
For a general comment, provided you can factor its denominator as a product of irreducible real or complex polynomials, there is a way to compute explicitely the antiderivative of a rational function. Your favorite Calculus textbook has probably a section about this. –  Taladris Aug 28 '14 at 5:56
@Taladris Is there a multiplication rule for integration or something (not related to this)? –  Hello Aug 28 '14 at 5:59

$$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt$$

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:

$$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt=\int_0^{\frac{\pi}{4}}2 du=2\frac{\pi}{4}=\frac{\pi}{2}$$

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$$\int_0^{\frac{1}{2}} \frac{4}{4t^2+1}\ dt$$ Let $u=2t$, $$\frac{d}{dt}u=\frac{d}{dt}[2t]=2 \Rightarrow du = 2\ dt$$ $$\int_0^1 \frac{2}{u^2+1}\ du= 2 \int_0^1 \frac{1}{u^2+1}\ du=2\arctan u\bigg|_0^1$$ $$=2\arctan 1 -2\arctan 0=2\frac{\pi}{4}-0=\frac{\pi}{2}$$

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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 $\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$.

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