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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- $\begingroup$ Note that $\displaystyle \int \dfrac4{1+4t^2}\mathrm dt=2\int \dfrac 2{1+(2t)^2}\mathrm dt$ and recall that $\displaystyle \int \dfrac{u'}{1+u^2}=\ldots$. $\endgroup$Git Gud– Git Gud2014-08-27 23:06:37 +00:00Commented Aug 27, 2014 at 23:06
- $\begingroup$ You can factor out the 4 in the bottom and use the arctangent formula, or substitute u=2t and then use this formula. $\endgroup$user84413– user844132014-08-27 23:08:24 +00:00Commented Aug 27, 2014 at 23:08
- $\begingroup$ 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 $\endgroup$Hello– Hello2014-08-27 23:09:20 +00:00Commented Aug 27, 2014 at 23:09
- $\begingroup$ 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. $\endgroup$Taladris– Taladris2014-08-28 05:56:59 +00:00Commented Aug 28, 2014 at 5:56
- $\begingroup$ @Taladris Is there a multiplication rule for integration or something (not related to this)? $\endgroup$Hello– Hello2014-08-28 05:59:41 +00:00Commented Aug 28, 2014 at 5:59
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3 Answers
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$$\int_0^{\frac{1}{2}} \frac{4}{1+4t^2} dt$$
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 $\int\frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan\frac{x}{a}+C$.
