MX2 Integration Marathon (2 Viewers)

Directrix · Jan 30, 2021

I'm afraid I have to step in here.

What is 1+1

notme123 · Jan 30, 2021

Directrix said:
I'm afraid I have to step in here.

What is 1+1

I think you need uni level maths to do that.

notme123 · Jan 30, 2021

Qeru said:
Let $\cos{x}=\tan{u} \implies-\sin{x} dx=-\sec^2{u}du$

$I=\int \sqrt{\frac{1+\tan^2{u}}{(1-\tan^2{u})^2}} \times \sec^2{u}du \quad \text{since} \quad \sin{x}=\sqrt{1-\tan^2{u}}$

$I=\int \frac{\sec^3{u}}{1-\tan^2{u}} du$

$I=\int \frac{\frac{1}{\cos^3{u}}}{1-\frac{\sin^2u}{\cos^2{u}} du$

$I=\int \frac{1}{\cos{u}(\cos^2{u}-\sin^2{u})} du$

Let $s=\sin{u} \implies ds=-\cos{u} du \implies -\frac{1}{\cos{u}}ds=du$

$I=-\int \frac{1}{(1-s^2)(1-2s^2)} du$

$I=\int \frac{1}{(s-1)(s+1)(1-\sqrt{2}s)(1+\sqrt{2}s)} ds$

$I=\int \frac{1}{(s-1)(s+1)(1-\sqrt{2}s)(1+\sqrt{2}s)} ds$

$I=\int \frac{1}{2}\left(\frac{1}{s+1}-\frac{1}{s-1}\right)+\frac{2}{(1-\sqrt{2}s)(1+\sqrt{2}s)}ds$ Ill save the partial fractions working.

$I=\int \frac{1}{2}\left(\frac{1}{s+1}-\frac{1}{s-1}\right)+\left(\frac{1}{1-\sqrt{2}s}+\frac{1}{1+\sqrt{2}s}\right)ds$

$I=\frac{1}{2} \ln\left(\frac{s+1}{s-1}\right)+\frac{1}{\sqrt{2}} \ln {\left(\frac{1+\sqrt{2}s}{1-\sqrt{2}s}\right)} +C$

Drawing up a triangle we find: $s=\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}$

So:

$I=\frac{1}{2} \ln\left(\frac{\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}+1}{\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}-1}\right)+\frac{1}{\sqrt{2}} \ln {\left(\frac{1+\sqrt{2}\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}}{1-\sqrt{2}\frac{\cos{x}}{\sqrt{\cos^2{x}+1}}}\right)} +C$

I could have very easily made a mistake whether in latex or just trying to evaluate to integral so let me know if you find any

with the s sub why when differentiated sin is negative cos

Qeru · Jan 30, 2021

notme123 said:
with the s sub why when differentiated sin is negative cos

Pretend you didnt see that

. Also yes rookie mistake and fixed.

notme123 · Jan 30, 2021

Qeru said:
Pretend you didnt see that . Also yes rookie mistake and fixed.

It's ok it's amazing you even were able to get to that point. I saw another one but I think I may be wrong because you and vern got to the same point but also the first sub why is the differential of tan -sec squared
Dw you fixed it

Qeru · Jan 30, 2021

notme123 said:
It's ok it's amazing you even were able to get to that point. I saw another one but I think I may be wrong because you and vern got to the same point but also the first sub why is the differential of tan -sec squared
Dw you fixed it

Yeah honestly don't know what I was thinking, ironically the final answer is correct since the two negatives (which were both mistakes) cancel out.

Qeru · Jan 30, 2021

notme123 said:
Something's very wrong, I can feel it.
$\int{ \sqrt{\csc^2{x}+\cot^2{x}} dx =\int{ \frac{\sqrt{1+\cos^2(x)}}{\sin(x)}dx$
$u^2=1+\cos^2x$
$2udu=-2\cos(x)*\sin(x)dx$
$dx=-\frac{u}{\sin(x)*\cos(x)}du$
$-\int{\frac{u^2}{\sin^2(x)*\cos(x)}du$
$=-\int{\frac{u^2}{\sqrt{u^2-1}*(2-u^2)}du$
$w^2=u^2-1$
$du=\frac{w}{u}dw$
$-\int{\frac{u^2}{w*(1-w^2)}*\frac{w}{u}dw$
$=\int{\frac{u}{{(w^2-1)}}dw$

Yeah everything is good till the last step when you have to convert the u to a w so you will get a square root again.

notme123 · Jan 30, 2021

Qeru said:
Yeah honestly don't know what I was thinking, ironically the final answer is correct since the two negatives (which were both mistakes) cancel out.

You're invincible at maths. Even when you make mistakes it's not really a mistake and works out in your favor.

notme123 · Jan 30, 2021

Qeru said:
Yeah everything is good till the last step when you have to convert the u to a w so you will get a square root again.

i just gave up at the end cuz i knew it was gonna look ugly
actually i couldve subbed in w = tanv or something

Qeru · Jan 30, 2021

notme123 said:
i just gave up at the end cuz i knew it was gonna look ugly
actually i couldve subbed in w = tanv or something

Yep and you would have got what Vern and I got. The sub $\cos{x}=\tan{u}$ does your two subs in one.

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

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