MX2 Integration Marathon (10 Viewers)

Qeru · Jan 30, 2021

idkkdi said:
Just a question. The standard way to do this integral seems to look like a page of working at least.
Any short ways?

After looking at my working I found a big brain substitution:
$\cos{x}=\frac{u}{\sqrt{1-u^2}}$

Which gets you straight to:

$\int \frac{1}{(1-2u^2)(u^2-1)} du$ so you can go partial fractions straight away.

YonOra · Jan 30, 2021

Seriously is a pleasure to read through this thread. Should've done 4u

stupid_girl · Jan 31, 2021

It's not too hard to get the antiderivative in terms of elementary function...but the evil is in the substitution of upper and lower limits.

$\int_{\frac{\pi}{3}}^{\frac{2021\pi}{2}}\frac{1}{2-\cos x}dx$

vernburn · Jan 31, 2021

stupid_girl said:
It's not too hard to get the antiderivative in terms of elementary function...but the evil is in the substitution of upper and lower limits.
$\int_{\frac{\pi}{3}}^{\frac{2021\pi}{2}}\frac{1}{2-\cos x}dx$

Finding the antiderivative is very easy using a t-sub. I've shown a different way the antiderivative can be found below:
$\int\frac{1}{2-\cos x}dx$
$=\int\frac{1}{2-\left(2\cos^2\frac{x}{2}-1\right)}dx$
$=\int\frac{1}{3-2\cos^2\frac{x}{2}}dx$
$=\int\frac{\sec^2\frac{x}{2}}{3\sec^2\frac{x}{2}-2}dx$
$=\int\frac{\sec^2\frac{x}{2}}{3\tan^2\frac{x}{2}+1}dx$
$=\frac{2}{\sqrt3}\tan^{-1}\left(\sqrt3\tan\frac{x}{2}\right)+C$
Now to evaluate the definite integral. The integrand is $2\pi$ periodic (with segments ...0~ $2\pi$ , $2\pi$ ~ $4\pi$ ,...) and the antiderivative is $2\pi$ periodic (with segments ... $-\pi$ ~ $\pi$ , $\pi$ ~ $2\pi$ ...). Note that we cannot just plug in the bounds due to this periodicity. In fact, it is only "safe" to integrate within ... $-\pi$ ~ $\pi$ , $\pi$ ~ $2\pi$ ...
Thus we can split the bounds into three regions: $\frac{\pi}{3}$ to $\pi$ , $\pi$ to $1010\pi$ (which we will treat as 1009 lots of 0 to $\pi$ by symmetry) and $1010\pi$ to $\frac{2021\pi}{2}$ .

Plugging all this into the antiderivative yields:
$1009\left(\frac{\pi}{\sqrt3}-0\right)+\left(\frac{\pi}{\sqrt3}-\frac{\pi}{2\sqrt3}\right)+\left(\frac{2\pi}{3\sqrt3}-0\right) = \boxed{\frac{6061\pi}{6\sqrt3}}$
(This ignores all the shenanigans with limits going on.)

I hope this explains my thinking thoroughly enough - a lot of this is intuitive but hard to communicate.

stupid_girl · Jan 31, 2021

Periodicity is not the cause of this trouble. If you integrate cos x, then you get sin x which is periodic. However, the substitution still works as usual.

The cause of this trouble is a different constant of integration at different intervals.

Instead of +c for all real numbers, you actually have +c0 for -pi<=x<=pi, +c1 for pi<=x<=3pi, +c2 for 3pi<=x<=5pi, etc.

vernburn · Jan 31, 2021

stupid_girl said:
Periodicity is not the cause of this trouble. If you integrate cos x, then you get sin x which is periodic. However, the substitution still works as usual.

The cause of this trouble is a different constant of integration at different intervals.

Instead of +c for all real numbers, you actually have +c0 for -pi<=x<=pi, +c1 for pi<=x<=3pi, +c2 for 3pi<=x<=5pi, etc.

Ok, thanks for the insight!

CM_Tutor · Jan 31, 2021

stupid_girl said:
Periodicity is not the cause of this trouble. If you integrate cos x, then you get sin x which is periodic. However, the substitution still works as usual.

The cause of this trouble is a different constant of integration at different intervals.

Instead of +c for all real numbers, you actually have +c0 for -pi<=x<=pi, +c1 for pi<=x<=3pi, +c2 for 3pi<=x<=5pi, etc.

Actually, the function

$y=\frac{1}{2-\cos{x}}$

has a period of $2\pi$ and its domain and range are

$x\in\mathbb{R} \qquad \qquad y\in\left[\frac{1}{3}, 1\right]$

In other words,

$\int_{2n\pi}^{2(n+1)\pi}{\frac{dx}{2-\cos{x}}}=\int_{0}^{2\pi}{\frac{dx}{2-\cos{x}}} \qquad \text{for all $n\in\mathbb{Z}$}$

and, more generally,

$\int_{2n\pi}^{2n\pi+\alpha}{\frac{dx}{2-\cos{x}}}=\int_{0}^{\alpha}{\frac{dx}{2-\cos{x}}} \qquad \qquad \text{. . . . . . . . . (*)}$

It is also symmetric about $x=\pi$ , and so

$\int_{0}^{\pi}{\frac{dx}{2-\cos{x}}}=\int_{\pi}^{2\pi}{\frac{dx}{2-\cos{x}}} \qquad \qquad \text{. . . . . . . . . (**)}$

It thus follows that

$\int_{\frac{\pi}{3}}^{\frac{2021\pi}{2}}{\frac{dx}{2-\cos{x}}}$

$\begin{align*} &=\int_{\frac{\pi}{3}}^{2\pi}{\frac{dx}{2-\cos{x}}}+\int_{2\pi}^{4\pi}{\frac{dx}{2-\cos{x}}}+\int_{4\pi}^{6\pi}{\frac{dx}{2-\cos{x}}}+...+\int_{1008\pi}^{1010\pi}{\frac{dx}{2-\cos{x}}}+\int_{1010\pi}^{\frac{2021\pi}{2}}{\frac{dx}{2-\cos{x}}} \\ &=\int_{\frac{\pi}{3}}^{2\pi}{\frac{dx}{2-\cos{x}}}+504\int_{0}^{2\pi}{\frac{dx}{2-\cos{x}}}+\int_{1010\pi}^{\frac{2021\pi}{2}}{\frac{dx}{2-\cos{x}}} \qquad \text{using $\alpha=2\pi$ and $n\in\{1, 2, 3, ..., 504\}$ in (*)} \\ &=\int_{\frac{\pi}{3}}^{2\pi}{\frac{dx}{2-\cos{x}}}+504\int_{0}^{2\pi}{\frac{dx}{2-\cos{x}}}+\int_{0}^{\frac{\pi}{2}}{\frac{dx}{2-\cos{x}}} \qquad \text{using $\alpha=\frac{\pi}{2}$ and $n=505$ in (*)} \\ &=\int_{\frac{\pi}{3}}^{\pi}{\frac{dx}{2-\cos{x}}}+1009\int_{0}^{\pi}{\frac{dx}{2-\cos{x}}}+\int_{0}^{\frac{\pi}{2}}{\frac{dx}{2-\cos{x}}} \qquad \text{using (**)} \\ &= \left(\frac{\pi}{\sqrt{3}}-\frac{\pi}{2\sqrt{3}}\right)+\frac{1009\pi}{\sqrt{3}}+\frac{2\pi}{3\sqrt{3}} \\ &=\frac{6061\pi}{6\sqrt{3}} \end{align*}$

So, I conclude that @vernburn is correct assuming that the integral (which I didn't check) is correct.

The problem arises because the inverse tan function can only return an angle between $-\pi / 2$ and $\pi / 2$ .

tito981 · Feb 1, 2021

This is the 2021 integration marathon thread, as i have noticed users submitting integrals on previous year's threads.

I will start off:
1/(x+x^6)
sorry for bad formatting

Qeru · Feb 1, 2021

tito981 said:
This is the 2021 integration marathon thread, as i have noticed users submitting integrals on previous year's threads.

I will start off:
1/(x+x^6)
sorry for bad formatting

$\int \frac{1}{x+x^6}dx$

$=\int \frac{1}{x^6(\frac{1}{x^5}+1)} dx$

$=-\frac{1}{5}\ln \left| \frac{1}{x^5}+1 \right|+C$

OR

$\int \frac{1}{x+x^6}dx$

$=\int \frac{1+6x^5-6x^5}{x+x^6}dx$

$=\int \frac{1+6x^5}{x+x^6}-\frac{6x^4}{1+x^5}dx$

$=\ln |x+x^6|-\frac{6}{5}\ln \left|1+x^5\right| +C$

Qeru · Feb 1, 2021

$\int \frac{1}{(x+a)^\frac{8}{7}(x-b)^\frac{6}{7}} dx$ where a and b are constants

vernburn · Feb 1, 2021

Qeru said:
$\int \frac{1}{(x+a)^\frac{8}{7}(x-b)^\frac{6}{7}} dx$ where a and b are constants

$\int \frac{1}{(x+a)^\frac{8}{7}(x-b)^\frac{6}{7}} dx$
$=\int \frac{1}{\left(\frac{x+a}{x-b}\right)^\frac{8}{7}(x-b)^2} dx$
Let $u=\frac{x+a}{x-b}\implies du = -\frac{a+b}{(x-b)^2}dx$ yielding:
$=-\frac{1}{a+b}\int\frac{1}{u^{\frac87}}du$
$=-\frac{1}{a+b}\left(-\frac{7}{u^\frac17}\right)+C$
$=\frac{7}{a+b}\left(\frac{x-b}{x+a}\right)^\frac17+C$

YonOra · Feb 1, 2021

$\int x dx$
Show all necessary working, please. V tricky.

Qeru · Feb 1, 2021

YonOra said:
$\int x dx$
Show all necessary working, please. V tricky.

Let $x=\sin{u} \implies dx=\cos{u} du$

$\int x dx$

$=\int \sin{u} \cos{u} du$

$=\frac{1}{2}\int \sin{2u} du$

$=-\frac{1}{4} \cos{2u}+C$ where C is a constant

$=-\frac{1}{4} (1-2\sin^2{x})+C$

$=-\frac{1}{4} (1-2x^2)+C$

$=-\frac{1}{4} +\frac{1}{2}x^2+C$

$=\frac{1}{2}x^2+C_1$ where $C_1$ is another constant

vernburn · Feb 1, 2021

Try:
$\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx$

YonOra · Feb 1, 2021

Qeru said:
Let $x=\sin{u} \implies dx=\cos{u} du$

$\int x dx$

$=\int \sin{u} \cos{u} du$

$=\frac{1}{2}\int \sin{2u} du$

$=-\frac{1}{4} \cos{2u}+C$ where C is a constant

$=-\frac{1}{4} (1-2\sin^2{x})+C$

$=-\frac{1}{4} (1-2x^2)+C$

$=-\frac{1}{4} +\frac{1}{2}x^2+C$

$=\frac{1}{2}x^2+C_1$ where $C_1$ is another constant

Couldn't have asked for more

notme123 · Feb 1, 2021

vernburn said:
Try:
$\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx$

$\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx$
$=\int\frac{x^2\tan^{-1}x}{1+x^2}dx$
$u=\arctan(x)$
$dx=(1+x^2) du$
$\int\(x^2)udu$
$= \int u(\tan^2(u))du$
$= \int u(\sec^2(u)-1)du$
$= \int u\sec^2(u)-udu$
$=u\tan(u) - \int\tan(u)du-\frac{u^2}{2} +C$
$=u\tan(u) -\ln|\sec(u)|-\frac{\arctan(x)^2}{2}+C$
$=x\arctan(x) -\ln|\sqrt{1+x^2}|-\frac{\arctan(x)^2}{2}+C$
$=x\arctan(x) -\frac{1}{2}\ln|1+x^2|-\frac{\arctan(x)^2}{2}+C$

Qeru · Feb 1, 2021

vernburn said:
Try:
$\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx$

Same adding and subtracting the same thing trick as the previous question:

$\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx$

$=\int\frac{x^2\tan^{-1}x}{1+x^2}dx$

$=\int\frac{(x^2+1-1)\tan^{-1}x}{1+x^2}dx$

$=\int \tan^{-1}(x)-\frac{\tan^{-1}x}{1+x^2}dx$

IBP on the first integral gives:

$I=x\tan^{-1}(x)-\int \frac{x}{x^2+1} dx=x\tan^{-1}(x)-\frac{1}{2}\ln(x^2+1)+C$

Second integral is just reverse chain rule

So in total: $\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx=x\tan^{-1}(x)-\frac{1}{2}\ln(x^2+1)-\frac{1}{2} (\tan^{-1}(x))^2+C_1$

idkkdi · Feb 1, 2021

vernburn said:
Try:
$\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx$

let x = tan u
x/ sec^2u / x^2 === > tan^3 u ===> tanusec^2u - tanu let tan u = z, z^2/2 + ln |cos u| lowkey messed it up, say where plz

notme123 · Feb 1, 2021

Qeru said:
Same adding and subtracting the same thing trick as the previous question:

$\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx$

$=\int\frac{x^2\tan^{-1}x}{1+x^2}dx$

$=\int\frac{(x^2+1-1)\tan^{-1}x}{1+x^2}dx$

$=\int \tan^{-1}(x)-\frac{\tan^{-1}x}{1+x^2}dx$

IBP on the first integral gives:

$I=x\tan^{-1}(x)-\int \frac{x}{x^2+1} dx=x\tan^{-1}(x)-\frac{1}{2}\ln(x^2+1)+C$

Second integral is just reverse chain rule

So in total: $\int\frac{\tan^{-1}x}{1+\frac{1}{x^2}}dx=x\tan^{-1}(x)-\frac{1}{2}\ln(x^2+1)-\frac{1}{2} (\tan^{-1}(x))^2+C_1$

this method is more intuitive and quicker I reckon esp if you know what the integral of inverse tan is off by heart.

vernburn · Feb 1, 2021

notme123 said:
this method is more intuitive and quicker I reckon esp if you know what the integral of inverse tan is off by heart.

This was the intended method.

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