Help with integration

**shubhadeep** · September 1st 2006, 06:31 AM

Help me with these integrations please

$ 1.\int\sqrt{log{x}}dx $
$ 2.\int\frac{dx}{\sqrt{3+2cos{x}}} $
$ 3.\int\log{tan(x)}dx $
$ 4.\int\frac{tan^{-1}(x)}{x}dx $
$ 5.\int\tan(e^{x})dx $
$ 6.\int\tan^{-1}(e^{x})dx $
$ 7.\int\(e^{(x^{2})}dx $

& the next one is a differential equation. Find the solution to the differential equation
$ (xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy=0 $

Integrals 3to7 were made up by me. Just wondering if indefinite integration is possible for them or not. The rest of them are book questions.

**ThePerfectHacker** · September 1st 2006, 07:06 AM

$ 7.\int\(e^{(x^{2})}dx $

This is not elementary.

Loiuville shown that if $f(x),g(x)$ are rational function then the integral $\int f(x) \exp (g(x)) dx$ is elementary if and only if, there exists a rational function $h(x)$ such as $f(x)=h'(x)+h(x)g(x)$ .

In this case, let us we have that.
$f(x)=1$ and $g(x)=x^2$ solve the differencial equation,
$1=y'+x^2y$ .
This equation has no rational solutions.
Thus the function is not elementary.
(However, it does exists. This is a consequence of the fundamental theorem of calculus. It can be shown that any countinous funtion on a closed interval or on the number line always has a an anti-derivative.)

If you are curious about which integral can and cannot be there is a topic called Differencial Algebra
---
Furthermore, this is an important function. It is used as an approximation of the binomial distribution (normal curve). I belive (I might be wrong) it is reffered to as $\mbox{erf}(x)$
---
Lasty, there is a nice idenity involving this function,
$\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$ .
If you wish I can post a prove of this (you need to be familiar with double integration).

**ThePerfectHacker** · September 1st 2006, 07:25 AM

Originally Posted by shubhadeep

$ 1.\int\sqrt{log{x}}dx $

Play around with it,
$\int \frac{x\sqrt{\log x}}{x} dx$
Let, $u=\log x$
Then, $u'=1/x$
Thus,
$\int e^u \sqrt{u} u' dx$
The substitution rule states,
$\int e^u \sqrt{u} du$
Manipulate it as,
$2\int e^{(\sqrt{u})^2} \frac{u}{2\sqrt{u}} du$
Let,
$t=\sqrt{u}$
Then,
$t'=\frac{1}{2\sqrt{u}}$
Thus,
$2\int e^{t^2} t^2 t' du$
Substitution rule again,
$2\int t^2 e^{t^2} dt$
Same idea not elementary.

**CaptainBlack** · September 1st 2006, 09:08 AM

Originally Posted by ThePerfectHacker

This is not elementary.

Loiuville shown that if $f(x),g(x)$ are rational function then the integral $\int f(x) \exp (g(x)) dx$ is elementary if and only if, there exists a rational function $h(x)$ such as $f(x)=h'(x)+h(x)g(x)$ .

In this case, let us we have that.
$f(x)=1$ and $g(x)=x^2$ solve the differencial equation,
$1=y'+x^2y$ .
This equation has no rational solutions.

Don't you have to show that this is the case?

RonL

**CaptainBlack** · September 1st 2006, 09:24 AM

Originally Posted by ThePerfectHacker

$ 7.\int\(e^{(x^{2})}dx $

This is not elementary.

But who said it had to be elementary? If we are looking for this
integral we usualy end up looking it up in a table, which tells
us that:

$ \int_0^x e^{x^2}=\frac{\sqrt{\pi}}{2}\mbox{erfi}(x) $

where $\mbox{erfi}(x)$ is the imaginary error function:

$\mbox{erfi}(z)=\bold{i}\ \mbox{erf}(\bold{i}z)$ , and $\mbox{erf}$ is the (analytic continuation of the) bog standard
error function:

$ \mbox{erf}(x)=\frac{2}{\sqrt{\pi}}\int_0^x e^{-t^2}\ dt $

RonL

**ThePerfectHacker** · September 1st 2006, 10:15 AM

Originally Posted by CaptainBlank

Don't you have to show that this is the case?

Do not understand. (You mean to show that the differencial equation has no rational functional solution? If thus, assume a solution exists as a quotient of two polynomials and try to reach a contradiction).

Originally Posted by CaptainBlank

But who said it had to be elementary?

Again do not understand. If you look at post #1 the user asks whether there exists closed form solutions. Thus, he did not know and I was saying negative.

**Soroban** · September 1st 2006, 10:18 AM

Hello, shubhadeep!

I took a stab at #1 . . . and it turned into #7 !

$1)\;\int\sqrt{\ln x}\,dx$

Let: $\sqrt{\ln x} = w\quad\Rightarrow\quad \ln x = w^2\quad\Rightarrow\quad x =$ $e^{w^2}\quad\Rightarrow\quad dx = 2w\!\cdot\! e^{w^2}dw$

Substitute: . $\int w\left(2we^{w^2}\,dw\right) \;=\;\int 2w^2\!\cdot\! e^{w^2}dw$

Integrate by parts . . .

. . Let: $u = w\qquad\quad dv = 2w\!\cdot\! e^{w^2}dw$

. . Then: $du = dw\qquad v = e^{w^2}$

and we have: . $w\!\cdot\!e^{w^2} - \int e^{w^2}dw$ . . . see?

**ThePerfectHacker** · September 1st 2006, 10:27 AM

Originally Posted by shubhadeep

$ 2.\int\frac{dx}{\sqrt{3+2cos{x}}} $

For this integral I recommend to use a Weierstrauss Substitution. It will eliminate the sine and transform the transcendental function into an algebraic one. This should make it easier to work with.

**CaptainBlack** · September 1st 2006, 10:48 AM

Originally Posted by ThePerfectHacker

Do not understand. (You mean to show that the differencial equation has no rational functional solution? If thus, assume a solution exists as a quotient of two polynomials and try to reach a contradiction).

I know how to approach it, but your reader may not, so you should
show it to be the case to complete the demonstration.

Again do not understand. If you look at post #1 the user asks whether there exists closed form solutions. Thus, he did not know and I was saying negative.

A closed form solution does not have to be closed form in terms of
elementary functions, it can be closed form in terms of higher
transcendental functions - that's what they are for.

RonL

**shubhadeep** · September 2nd 2006, 07:33 PM

What does being elementary mean. I'm only a beginner.

Originally Posted by ThePerfectHacker

For this integral I recommend to use a Weierstrauss Substitution. It will eliminate the sine and transform the transcendental function into an algebraic one. This should make it easier to work with.

Please be a little more descriptive. Again, I'm just a beginner.

Originally Posted by Soroban

Hello, shubhadeep!

I took a stab at #1 . . . and it turned into #7 !

Let: $\sqrt{\ln x} = w\quad\Rightarrow\quad \ln x = w^2\quad\Rightarrow\quad x =$ $e^{w^2}\quad\Rightarrow\quad dx = 2w\!\cdot\! e^{w^2}dw$

Substitute: . $\int w\left(2we^{w^2}\,dw\right) \;=\;\int 2w^2\!\cdot\! e^{w^2}dw$

Integrate by parts . . .

. . Let: $u = w\qquad\quad dv = 2w\!\cdot\! e^{w^2}dw$

. . Then: $du = dw\qquad v = e^{w^2}$

and we have: . $w\!\cdot\!e^{w^2} - \int e^{w^2}dw$ . . . see?

Yeah, I arrived at the same situation. That is why I posted 7 and many others which I arrived at while solving others.

**ThePerfectHacker** · September 3rd 2006, 05:26 AM

Originally Posted by shubhadeep

What does being elementary mean?

It means a finite combination of function addition,subtraction,multiplication,division and composition of the "basic functions" (ploynomials, rationals, power, exponenets, lgarithmic, trigonometric).

I myself never liked that definition because it seems to be missing parts to it. Never like when we say a finite composition.
---
This is my own question. Is there a more rigourous way to define elementary? Like a Group that is solvable, or something like that?

**ThePerfectHacker** · September 3rd 2006, 06:08 AM

Originally Posted by shubhadeep

Please be a lot more descriptive.

I assumed you understood.
---
You have,
$\int \frac{dx}{\sqrt{3+2\cos x}}$
Let,
$u=\tan x/2$
Then,
$\sin x=2\sin x/2\cos x/2=\frac{2u}{1+u^2}$
And,
$\cos x=\cos^2 x/2-\sin^2 x/2=\frac{1-u^2}{1+u^2}$
If you substitute this you will get,
$\int \frac{\frac{2}{1+u^2}}{\sqrt{3+2\frac{2u}{1+u^2}}} du$

**ThePerfectHacker** · September 3rd 2006, 07:35 AM

Originally Posted by galactus

$(xy^{2}-\frac{e}{x^{3}})dx-x^{2}ydy$ is exact since

$\frac{\partial{M}}{\partial{y}}=2xy=\frac{\partial {N}}{\partial{x}}$

How?
$\frac{\partial M}{\partial y}=2xy$
But!
$\frac{\partial N}{\partial x}=-2xy$
They are not equal.

**galactus** · September 3rd 2006, 08:16 AM

You're so right PH. I got tripped up with a mere negative sign. Oh well.

BTW, I ran $\int\frac{1}{\sqrt{3+2cos(x)}}dx$ through Maple.

The contrary LaTex on this site won't allow me to post the lengthy solution, but here's what it will post. At least, it'll give you an idea that it's a dilly.

Elliptic Fresnel means it's a real booger to integrate(by elementary means, anyway)

Here's what she gave me:

${-}2{\frac {\sqrt {- \left( 1+4\, \left( \cos \left( 1/2\,x \right)$
$\right) ^{2} \right) \left( -1+ \left( \cos \left( 1/2\,x \right)$
$\right) ^{2} \right) }\sqrt {1- \left( \cos \left( 1/2\,x \right)$
$\right) ^{2}}{\it EllipticF}$ $\left( \cos \left( 1/2\,x \right)$ $,2\,i \right) }{\sqrt {1-4\, \left( \cos \left( 1/2\,x \right) \right)^{4}+3$ $\, \left( \cos \left( 1/2\,x \right) \right) ^{2}}\sin \left( 1/2\,x \right) }}$

**shubhadeep** · September 3rd 2006, 09:39 AM

Originally Posted by ThePerfectHacker

I assumed you understood.

Hmm...I'm familiar with this type of substitution but never came to know its called Weierstrauss Substitution.

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