Finding Infinite Areas are Awesome!

From this question, what we are being asked is to find

$\displaystyle \int_{-\infty}^{\infty} \dfrac {360}{36 + x^2} \mathrm{d} x$

We can split this into

$\displaystyle \int_{-\infty}^{0} \dfrac {360}{36 + x^2} \mathrm{d} x + \displaystyle \int_{0}^{\infty} \dfrac {360}{36 + x^2} \mathrm{d} x$

From this, we can change both these integrals into

$\displaystyle \lim_{n \rightarrow -\infty} \int_{n}^0 \dfrac {360}{36 + x^2} \mathrm{d} x+ \lim_{m \rightarrow \infty} \int_0^{m} \dfrac {360}{36 + x^2} \mathrm{d} x$

Now, let's work on the integrals.

$\displaystyle \int \dfrac {360}{36 + x^2} \mathrm{d} x$

Well, the denominator looks promising since it is in the form of $a^2 + b^2$ . In this case, $a = 6$ and $b = x$ . We can now try to substitute $x$ with a different value. For this, we will do

$x = 6 \tan \theta$

Now to make sure that we can substitute back, let's change the subject to $\theta$ .

$\dfrac {x}{6} = \tan \theta$

$\arctan ( \dfrac {x}{6}) = \theta$

We also need to find $\mathrm{d} x$ of $6 \tan \theta$ . Simplifying, we get

$6 \sec^2 \theta \mathrm{d} \theta$

Inputting these values back in the integral, we get

$\displaystyle \int \dfrac {360 \cdot 6 \sec^2 \theta}{36 + (6 \tan \theta)^2} \mathrm{d} \theta$

which is simplified as

$\displaystyle \int \dfrac {360 \cdot 6 \sec^2 \theta}{36 + 36 \tan^2 \theta} \mathrm{d} \theta$

The denominator can be factorised to

$\displaystyle \int \dfrac {360 \cdot 6 \sec^2 \theta}{36(1 + \tan^2 \theta} \mathrm{d} \theta$

which simplifies the fraction as

$\displaystyle \int \dfrac {10 \cdot 6 \sec^2 \theta}{1 + \tan^2 \theta} \mathrm{d} \theta$

Notice that $1 + \tan^2 \theta$ can be written as $\sec^2 \theta$ . With this in mind, we simplify further to just

$\displaystyle \int 60 \mathrm{d} \theta$

or

$60 \displaystyle \int \mathrm{d} \theta$

which is simply

$60 \theta + C$

Now we substitute the value of $x$ back in to get

$60 (\arctan \left (\dfrac {x}{6}) \right)$

or

$60 \arctan ( \dfrac {x}{6})$

Now we take this value back to the limits. Sustituting them, we get

$\displaystyle \lim_{n \rightarrow -\infty} (60 \arctan \left (\dfrac {x}{6}) \right)|_{n}^{0} + \lim_{m \rightarrow \infty} (60 \arctan \left (\dfrac {x}{6}) \right)|_{0}^{m}$

Now, if we simplify this, we get

$\displaystyle \lim_{n \rightarrow -\infty} (60 \arctan (\dfrac {0}{6}) - 60 \arctan (\dfrac {n}{6})) + \lim_{m \rightarrow \infty} (60 \arctan (\dfrac {m}{6}) - 60 \arctan (\dfrac {0}{6}))$

Well, $\arctan (0)$ is just 0. For $-\infty$ , it's $\frac {-\pi}{2}$ and for $\infty$ , it's $\frac {\pi}{2}$ . Inputting these values, we get

$(60 \cdot 0 - 60 \cdot \dfrac {-\pi}{2}) + (60 \cdot \dfrac {\pi}{2} - 60 \cdot 0)$

or just

$30 \pi + 30 \pi = 60 \pi$

Therefore, $60$ is our answer. Upvotes would be appreciated since I spent over an hour on this solution.

Consider the integral $\displaystyle\int \dfrac{1}{a^2+x^2}\text d x$ . Let $x=a\tan z$ so that $\text d x=a \sec^2 z$ . Putting these yields $\int \dfrac{a \sec^2 z}{a^2\left(1+\tan^2 z\right)}\text d z=\int \dfrac{\sec^2 z}{a \sec^2 z}\text d z=\int \dfrac 1 a\text d z=\dfrac{z}{a}+\text C=\dfrac{\arctan(x/a)}{a}+\text C.$ Notice that the function $y=\dfrac{1}{a^2+x^2}$ is even. So we have $\int_{-\infty}^\infty \dfrac{1}{a^2+x^2}\text d x=2\int_0^\infty \dfrac{1}{a^2+x^2}\text d x=2\left.\dfrac{\arctan(x/a)}{a}\right|_{0}^\infty=\dfrac{\pi}{a}.$ Now plugging in $a=6$ and multiplying by $360$ gives our desired answer $\dfrac{360\pi}{6}=60\pi$ .

I still have that "Table of Integrals" book with me since high school days, and this one is an oldie

$\displaystyle\int { \dfrac { 1 }{ { a }^{ 2 }+{ x }^{ 2 } } } dx=\dfrac { 1 }{ a } ArcTan\left( \dfrac { x }{ a } \right)$

I didn't evaluate this integral as you have here. Thanks for the demonstration.