MX2 Integration Marathon 2021 (1 Viewer)

CM_Tutor

Moderator
Moderator
more integrals

$\bg_white \int \frac{\sin{\sqrt{x}+a}(e^{\sqrt{x}})}{\sqrt{x}}$
Do you mean

\bg_white \begin{align*} \int \frac{e^{\sqrt{x}}\sin{(\sqrt{x}+a)}}{\sqrt{x}}\ dx &= 2\int{e^u\sin(u+a)\ du} \qquad \text{where u=\sqrt{x}} \\ &= e^{\sqrt{x}}\left[\sin(\sqrt{x}+a)-\cos(\sqrt{x}+a)\right] + C \end{align*}

idkkdi

Well-Known Member
Do you mean

\bg_white \begin{align*} \int \frac{e^{\sqrt{x}}\sin{(\sqrt{x}+a)}}{\sqrt{x}}\ dx &= 2\int{e^u\sin(u+a)\ du} \qquad \text{where u=\sqrt{x}} \\ &= e^{\sqrt{x}}\left[\sin(\sqrt{x}+a)-\cos(\sqrt{x}+a)\right] + C \end{align*}
yes

CM_Tutor

Moderator
Moderator
At first, I thought you meant a much more difficult problem:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx$

idkkdi

Well-Known Member
At first, I thought you meant a much more difficult problem:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx$
@Qeru here ya go m8, im out

Qeru

Well-Known Member
At first, I thought you meant a much more difficult problem:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx$
Surely theres no elementary anti-derivative to this

CM_Tutor

Moderator
Moderator
Surely theres no elementary anti-derivative to this
None that I am aware of.

CM_Tutor

Moderator
Moderator
Though I can give a 1 line answer:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx = F(x) \quad \text{where} \quad \frac{dF}{dx} = \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}$

idkkdi

Well-Known Member
Though I can give a 1 line answer:

$\bg_white \int \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}\ dx = F(x) \quad \text{where} \quad \frac{dF}{dx} = \frac{\sin$e^{\sqrt{x}}(\sqrt{x}+a)$}{\sqrt{x}}$
... that isnt the integral.

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

Qeru

Well-Known Member
... 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

Well-Known Member
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

Well-Known Member
Ok anyone want to try this (stumped me lol):

$\bg_white \int \frac{\sqrt{\sin{\sqrt{x}}}\cos{\sqrt{x}}}{1+x^2} dx$
Have you tried this @CM_Tutor I'm pretty sure theres an anti-derivative it's just really tricky.

CM_Tutor

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

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

CM_Tutor

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

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

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

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

and it cannot be expressed in elementary functions.

However, there is a closed form for this integral:

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

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

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

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

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

You will need something of the form:

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

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

\bg_white \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 $\bg_white J$ alone as it is clearly a problem:

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

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

\bg_white \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*}

s97127

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

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

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

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

and it cannot be expressed in elementary functions.

However, there is a closed form for this integral:

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

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

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

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

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

You will need something of the form:

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

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

\bg_white \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 $\bg_white J$ alone as it is clearly a problem:

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

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

\bg_white \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

Active Member
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.