Help w/ Trig Integrals

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• Jul 14th 2007, 02:03 AM
Help w/ Trig Integrals
$\displaystyle \int(\sin{x})^3(\cos{x})^3dx$
am i going to use trigonometric or int by parts?
help
• Jul 14th 2007, 04:34 AM
Soroban

I have to ask . . .
Are those really $\displaystyle x^3$ ?

We can integrate: $\displaystyle \sin^3\!x\cos^3\!x$ . . . but not $\displaystyle \sin(x^3)\cos(x^3)$

• Jul 14th 2007, 04:35 AM
oopss
its $\displaystyle (\sin{x})^3 (\cos{x})^3$
• Jul 14th 2007, 04:40 AM
topsquark
$\displaystyle \int sin^3(x) cos^3(x) dx$

Let $\displaystyle y = sin(x)$, then $\displaystyle dy = cos(x) dx$

$\displaystyle \int sin^3(x) cos^3(x) dx = \int sin^3(x) cos^2(x) cos(x) dx$

$\displaystyle = \int sin^3(x) (1 - sin^2(x)) cos(x) dx = \int y^3 (1 - y^2) dy$

$\displaystyle = \int (y^3 - y^5)$

I'm sure you can take it from here.

-Dan
• Jul 14th 2007, 04:47 AM
im sure that either sin(x) or cos(x) is integrable
but the integrator has a different answer
maybe use integration by parts?
• Jul 14th 2007, 06:37 AM
Jhevon
Quote:

Originally Posted by Soroban

We can integrate: $\displaystyle \sin^3\!x\cos^3\!x$ . . . but not $\displaystyle \sin(x^3)\cos(x^3)$

Can someone tell me once and for all how can you tell if you can't integrate something, or something is not integrable analytically using elementary functions or whatever...wait, did i ask this question before?

I know we can't integrate $\displaystyle e^{x^2}$ and according to Soroban, we can't integrate $\displaystyle \sin \left( x^3 \right) \cos \left( x^3 \right)$, but how do we know that for sure? What's the proof that we can't integrate those functions by hand?
• Jul 14th 2007, 07:13 AM
Plato
This totally my own opinion: Your confusion is understandable and it comes from the very sad conflating of the words integral and antiderivative. They are not the same. An integral is a number, quite often gotten by way of an antiderivative using the fundamental theorem of integral calculus. An antiderivative is just what is says it is.

Of the antiderivative of $\displaystyle \sin(x^3)$ does exits but we would the series representation for $\displaystyle \sin(x)$ to get it. Therefore, it is proper to say that no elementary representation of the antiderivative of $\displaystyle \sin(x^3)$ exist in the Calculus II sense of the term.
• Jul 14th 2007, 07:23 AM
topsquark
Quote:

Originally Posted by topsquark
$\displaystyle \int sin^3(x) cos^3(x) dx$

Let $\displaystyle y = sin(x)$, then $\displaystyle dy = cos(x) dx$

$\displaystyle \int sin^3(x) cos^3(x) dx = \int sin^3(x) cos^2(x) cos(x) dx$

$\displaystyle = \int sin^3(x) (1 - sin^2(x)) cos(x) dx = \int y^3 (1 - y^2) dy$

Now,

$\displaystyle = \int (y^3 - y^5)$

I'm sure you can take it from here.

-Dan

To continue
$\displaystyle = \frac{1}{4}y^4 - \frac{1}{6}y^6 + C$

$\displaystyle = \frac{1}{4}sin^4(x) - \frac{1}{6}sin^6(x) + C$

Now, my TI-92 comes up with:
$\displaystyle -\frac{sin^2(x) cos^4(x)}{6} - \frac{cos^4(x)}{12}$

and the Integrator comes up with
$\displaystyle \frac{1}{192} ( cos(6x) - 9 cos(2x))$

All of these solutions are correct, despite how it might look. The point is that we are doing indefinite integration, so any solution that differs from another by only a constant are all correct. If you spend the time (or just plug it through on your calculator) you will find that all three solutions differ from each other by some constant. (Neither the TI-92 nor the Integrator remind you to add the arbitrary constant on the end.)

-Dan
• Jul 14th 2007, 07:27 AM
thanks topsquark
ohh ok
• Jul 14th 2007, 07:43 AM
Krizalid
You can use the fact

$\displaystyle \int\sin^mx\cos^nx~dx={\color{blue}\frac{\sin^{m+1 }(x)\cos^{n-1}(x)}{m+n}+\frac{n-1}{m+n}\int\sin^mx\cos^{n-2}(x)~dx},~m\ne-n$

:D:D
• Jul 14th 2007, 09:12 AM
galactus
Quote:

Originally Posted by Jhevon
Can someone tell me once and for all how can you tell if you can't integrate something, or something is not integrable analytically using elementary functions or whatever...wait, did i ask this question before?

I know we can't integrate $\displaystyle e^{x^2}$ and according to Soroban, we can't integrate $\displaystyle \sin \left( x^3 \right) \cos \left( x^3 \right)$, but how do we know that for sure? What's the proof that we can't integrate those functions by hand?

I believe sin(x^3) is done using what is known as a Lommel integral. Don't know much about it though. Just as sin(x^2) is a Fresnel.

I couldn't find reference to Lommel in wiki. Perhaps, that would be a good MathHelpWiki for someone to take on?.

One should be able to use topics from advanced calc to prove sin(x^3) in not integrable by elementary means. Maybe Dirichlet test or something.

It is continuous and differentiable.

I may have to delve into it some more.
• Jul 14th 2007, 10:13 AM
DivideBy0
As far as I know,

$\displaystyle \int {\sin x^3 ~dx} = - \frac{1} {2}i\left( {\frac{{x\Gamma\left( {\displaystyle\frac{1} {3},ix^3 } \right)}} {{3\sqrt[3]{{ix^3 }}}} - \frac{{x\Gamma\left( {\displaystyle\frac{1} {3}, - ix^3 } \right)}} {{3\sqrt[3]{{ - ix^3 }}}}} \right) + k$
• Jul 14th 2007, 02:46 PM
galactus
Yeah. I ran it through Maple and it gave me a horrendous result with LommelSi. It may be equivalent to your result, though. Just a different animal.
• Jul 14th 2007, 04:03 PM
Krizalid
Quote:

Originally Posted by DivideBy0
As far as I know,

$\displaystyle \int {\sin x^3 ~dx} = - \frac{1} {2}i\left( {\frac{{x\Gamma\left( {\displaystyle\frac{1} {3},ix^3 } \right)}} {{3\sqrt[3]{{ix^3 }}}} - \frac{{x\Gamma\left( {\displaystyle\frac{1} {3}, - ix^3 } \right)}} {{3\sqrt[3]{{ - ix^3 }}}}} \right) + k$

I know the forum where you got that :D:D
• Jul 14th 2007, 05:42 PM
ThePerfectHacker
I can find,
$\displaystyle \int_0^{\infty} \sin x^3 \cos x^3 dx$ :cool:

And,
$\displaystyle \int_0^{\infty} \sin x^3 dx$
And,
$\displaystyle \int_0^{\infty} \cos x^3 dx$

Eventhough these functions are not elementary.
(I sometimes love Complex Analysis).
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