MX2 Integration Marathon (1 Viewer)

Qeru · Feb 2, 2021

idkkdi said:
... that isnt the integral.

am i missing something? this doesn't look like a one line answer.

He basically defined that integral as a function similar to the error function or Si(x) etc.

idkkdi · Feb 2, 2021

Qeru said:
He basically defined that integral as a function similar to the error function or Si(x) etc.

what. it looks like he diffed the integral to get what's the integrand to me.

Qeru · Feb 2, 2021

Qeru said:
Ok anyone want to try this (stumped me lol):

$\int \frac{\sqrt{\sin{\sqrt{x}}}\cos{\sqrt{x}}}{1+x^2} dx$

Have you tried this @CM_Tutor I'm pretty sure theres an anti-derivative it's just really tricky.

s97127 · Feb 2, 2021

CM_Tutor · Feb 2, 2021

Qeru said:
Ok anyone want to try this (stumped me lol):

$\int \frac{\sqrt{\sin{\sqrt{x}}}\cos{\sqrt{x}}}{1+x^2} dx$

I don't see an obvious way to approach this. Integral calculator can't find an answer in elementary functions, which suggests that none of the common approaches will work. What makes you confident that there is an answer with the realm of MX2 possibilities?

Qeru · Feb 2, 2021

CM_Tutor said:
I don't see an obvious way to approach this. Integral calculator can't find an answer in elementary functions, which suggests that none of the common approaches will work. What makes you confident that there is an answer with the realm of MX2 possibilities?

I found it on this other forum for Calc 1 (forgot where). Also I found this hint: https://math.stackexchange.com/ques...he-integral-of-sqrt-sin-sqrt-x-cos-sqrt-x-1x2

Not sure if that leads anywhere though

CM_Tutor · Feb 2, 2021

s97127 said:
View attachment 30180

This is much tougher than I first thought as you keep running into the exponential integral, $\text{Ei}(x)$ (or a transform of it), being integrals such as

$\int{\frac{e^x}{x}}\ dx$

Transforms include the denominator being a linear function of $x$ or the numerator as $e^{-x}$ . Properly, the exponential integral is defined (over the complex plane) as

$\text{Ei}(x)=\int_{-x}^{\infty}{\frac{e^{-t}}{t}}\ dt=\int_{-\infty}^{x}{\frac{e^t}{t}}\ dt$

and it cannot be expressed in elementary functions.

However, there is a closed form for this integral:

$\int{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx =\frac{xe^x}{x+2} +C \text{, for some constant $C$}$

$\int_0^1{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx =\frac{e}{3}$

For those who want to try to solve the problem yourselves, you need to rewrite the integrand without creating terms of the form

$\frac{Ae^x}{(mx+b)^n$

where $A$ , $m$ , and $b$ , are constants and, in this case, $n=1$ or $n=2$ .

You will need something of the form:

$\left(\frac{f(x)}{x+2}+\frac{g(x)}{(x+2)^2}\right)e^x$

where $f(x)$ and $g(x)$ are both of degree 1.

$\begin{align*} I&=\int_0^1{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx \\ &=\int_0^1{\frac{x^2+3x + 2 - x}{(x+2)^2}e^x}\ dx \\ &=\int_0^1{\left(\frac{(x+1)(x+2) - x}{(x+2)^2}\right)e^x}\ dx \\ &=\int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx - \int_0^1{\frac{xe^x}{(x+2)^2}}\ dx \\ &=J - \int_0^1{\frac{xe^x}{(x+2)^2}}\ dx \qquad \text{where } J = \int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx \end{align*}$

We know to leave $J$ alone as it is clearly a problem:

$J = \int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx = \int_0^1{\frac{(x + 2 - 1)e^x}{x+2}}\ dx = \int_0^1{e^x - \frac{e^x}{x+2}}\ dx$

i.e. $J$ is a constant plus (or minus) the exponential integral.

$\begin{align*} I&=J - \int_0^1{xe^x}\ d\left(\frac{-1}{x+2}\right) \\ &=J - \left(\left[\frac{-xe^x}{x+2}\right]_0^1 + \int_0^1{\frac{-1}{x+2}}\ d(xe^x)\right) \\ &=J - \left[\frac{-1 \times e^1}{1+2} - \frac{-0 \times e^0}{0+2}\right] - \int_0^1{\frac{e^x + xe^x}{x+2}}\ dx \\ &=J + \frac{e}{3} - 0 - \int_0^1{\frac{e^x(x+1)}{x+2}}\ dx \\ &=J + \frac{e}{3} - J \\ &= \frac{e}{3} \end{align*}$

CM_Tutor · Feb 2, 2021

Qeru said:
I found it on this other forum for Calc 1 (forgot where). Also I found this hint: https://math.stackexchange.com/ques...he-integral-of-sqrt-sin-sqrt-x-cos-sqrt-x-1x2

Not sure if that leads anywhere though

I found the transformations that are suggested there, and others, but none seems to lead in a productive direction.

s97127 · Feb 2, 2021

CM_Tutor said:
This is much tougher than I first thought as you keep running into the exponential integral, $\text{Ei}(x)$ (or a transform of it), being integrals such as

$\int{\frac{e^x}{x}}\ dx$

Transforms include the denominator being a linear function of $x$ or the numerator as $e^{-x}$ . Properly, the exponential integral is defined (over the complex plane) as

$\text{Ei}(x)=\int_{-x}^{\infty}{\frac{e^{-t}}{t}}\ dt=\int_{-\infty}^{x}{\frac{e^t}{t}}\ dt$

and it cannot be expressed in elementary functions.

However, there is a closed form for this integral:

$\int{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx =\frac{xe^x}{x+2} +C \text{, for some constant $C$}$

$\int_0^1{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx =\frac{e}{3}$

For those who want to try to solve the problem yourselves, you need to rewrite the integrand without creating terms of the form

$\frac{Ae^x}{(mx+b)^n$

where $A$ , $m$ , and $b$ , are constants and, in this case, $n=1$ or $n=2$ .

You will need something of the form:

$\left(\frac{f(x)}{x+2}+\frac{g(x)}{(x+2)^2}\right)e^x$

where $f(x)$ and $g(x)$ are both of degree 1.

$\begin{align*} I&=\int_0^1{\frac{x^2+2x+2}{x^2+4x+4}e^x}\ dx \\ &=\int_0^1{\frac{x^2+3x + 2 - x}{(x+2)^2}e^x}\ dx \\ &=\int_0^1{\left(\frac{(x+1)(x+2) - x}{(x+2)^2}\right)e^x}\ dx \\ &=\int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx - \int_0^1{\frac{xe^x}{(x+2)^2}}\ dx \\ &=J - \int_0^1{\frac{xe^x}{(x+2)^2}}\ dx \qquad \text{where } J = \int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx \end{align*}$

We know to leave $J$ alone as it is clearly a problem:

$J = \int_0^1{\frac{(x + 1)e^x}{x+2}}\ dx = \int_0^1{\frac{(x + 2 - 1)e^x}{x+2}}\ dx = \int_0^1{e^x - \frac{e^x}{x+2}}\ dx$

i.e. $J$ is a constant plus (or minus) the exponential integral.

$\begin{align*} I&=J - \int_0^1{xe^x}\ d\left(\frac{-1}{x+2}\right) \\ &=J - \left(\left[\frac{-xe^x}{x+2}\right]_0^1 + \int_0^1{\frac{-1}{x+2}}\ d(xe^x)\right) \\ &=J - \left[\frac{-1 \times e^1}{1+2} - \frac{-0 \times e^0}{0+2}\right] - \int_0^1{\frac{e^x + xe^x}{x+2}}\ dx \\ &=J + \frac{e}{3} - 0 - \int_0^1{\frac{e^x(x+1)}{x+2}}\ dx \\ &=J + \frac{e}{3} - J \\ &= \frac{e}{3} \end{align*}$

That's correct. what's the hardest integration problem that you have solved?

stupid_girl · Feb 2, 2021

I know @vernburn has correctly split the interval. However, periodicity actually made the calculation easier rather than causing trouble.

Depending on how you express the antiderivative, the constant of integration may be different.

stupid_girl · Feb 2, 2021

I made a typo in Wolfram earlier. It does give a nice answer.
$\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$

vernburn · Feb 2, 2021

stupid_girl said:
Not sure why Wolfram can't give a nice answer.
$\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$

Let $u = -x \implies du = -dx$ :
$I =\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}dx$
$=-\int_{1}^{-1} \frac{(-u-1)e^{-u}}{u^2+e^{-2u}}du$
$=\int_{1}^{-1}\frac{(u+1)e^u}{e^{2u}u^2+1}du$
Notice that $(u+1)e^u$ is the derivative of $ue^u$ and hence the integral is in the form: $\int \frac{f'(x)}{\left[f(x)\right]^2+1}dx = \tan^{-1}f(x)+C$ . Thus,
$I = \left[\tan^{-1}\left(ue^u\right)\right]_{1}^{-1}$
$=-\tan^{-1}\left(\frac{1}{e}\right)-\tan^{-1}(e)$
$=\boxed{-\frac{\pi}{2}}$ (using the well-known identity $\tan^{-1}x+\tan^{-1}\frac{1}{x} = \frac{\pi}{2}$ )

s97127 · Feb 2, 2021

vernburn said:
Let $u = -x \implies du = -dx$ :
$I =\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}dx$
$=-\int_{1}^{-1} \frac{(-u-1)e^{-u}}{u^2+e^{-2u}}du$
$=\int_{1}^{-1}\frac{(u+1)e^u}{e^{2u}u^2+1}du$
Notice that $(u+1)e^u$ is the derivative of $ue^u$ and hence the integral is in the form: $\int \frac{f'(x)}{\left[f(x)\right]^2+1}dx = \tan^{-1}f(x)+C$ . Thus,
$I = \left[\tan^{-1}\left(ue^u\right)\right]_{1}^{-1}$
$=-\tan^{-1}\left(\frac{1}{e}\right)-\tan^{-1}(e)$
$=\boxed{-\frac{\pi}{2}}$ (using the well-known identity $\tan^{-1}x+\tan^{-1}\frac{1}{x} = \frac{\pi}{2}$ )

how come a medical student like yourself still has so much interest in math after finishing HSC?

tito981 · Feb 2, 2021

s97127 said:
how come a medical student like yourself still has so much interest in math after finishing HSC?

once u learn math properly u never forget.

CM_Tutor · Feb 3, 2021

stupid_girl said:
I made a typo in Wolfram earlier. It does give a nice answer.
$\int_{-1}^{1} \frac{(x-1)e^x}{x^2+e^{2x}}$

Interesting... integration calculator doesn't

idkkdi · Feb 3, 2021

s97127 said:
how come a medical student like yourself still has so much interest in math after finishing HSC?

I have a feeling @vernburn likes maths more than the stuff he is going to do in med.

YonOra · Feb 3, 2021

Med is science, so how could u not get bored

Qeru · Feb 3, 2021

The ultimate tedious integral:

$\int \sqrt[5]{\tan{x}} dx$

CoolKids101 · Feb 3, 2021

Qeru said:
The ultimate tedious integral:

$\int \sqrt[5]{\tan{x}} dx$

You should attend the integration bee that's being held at UNSW lol

vernburn · Feb 4, 2021

idkkdi said:
I have a feeling @vernburn likes maths more than the stuff he is going to do in med.

It does look like it doesn’t it! Jkjk
In reality, I really only like integration now because there is a certain elegance to it and skill required (and it’s still fresh in my head). I find the rest of maths quite boring and stale imho. My attempts at the above integrals are really just due to boredom (not long till uni starts now though).

Qeru said:
The ultimate tedious integral:

$\int \sqrt[5]{\tan{x}} dx$

I may be bored but not enough to attempt this monster!

MX2 Integration Marathon (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Moderator

Well-Known Member

Moderator

Moderator

Active Member

Active Member

Active Member

Active Member

Active Member

Well-Known Member

Moderator

Well-Known Member

Well-Known Member

Well-Known Member

New Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)