Ingtegral

**Mathstud28** · April 16th 2008, 04:17 PM

ok so where did I go wrong.. $\int{cos(mx^2)^2dx}$ ...

so I first did this $\int{cos(mx^2)^2dx}=\int{\frac{1+cos(2mx^2)}{2}dx}$ ...so then I seperated it to get

$\int{\bigg[\frac{1}{2}+\frac{cos(2mx^2)}{2}dx\bigg]}$ $\Rightarrow{\int{\frac{1}{2}dx}+\int{\frac{cos(2mx ^2)}{2}dx}\Rightarrow{\frac{x}{2}+\int{\frac{cos(2 mx^2)}{2}dx}}}$

Then using substitution by power series I got

$\frac{x}{2}+\int\frac{\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(2mx^2)^{2n}}{(2n)!}}{2}dx\Rightarrow\fra c{x}{2}+\int\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{m^{2n}}\cdot{2^{2n-1}}\cdot{x^{4n}}}{(2n)!}dx$

...then I went

$\frac{x}{2}+\sum_{n=0}^{\infty}\frac{(-1)^{n}\cdot{m^{2n}}\cdot{2^{2n-1}}\cdot{x^{4n+1}}}{(4n+1)(2n)!}$

...could someone point out my mistake...thanks

**Mathstud28** · April 16th 2008, 05:18 PM

O and by the way this is for leisurely purposes so no rush

**Moo** · April 16th 2008, 11:35 PM

Hello,

$\frac{x}{2}+\int\frac{\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(2mx^2)^{\color{red}2n}}{(2n)!}}{2}dx$

Huh ?

**Mathstud28** · April 17th 2008, 02:36 AM

Originally Posted by Moo

Hello,

$\frac{x}{2}+\int\frac{\sum_{n=0}^{\infty}\frac{(-1)^n\cdot(2mx^2)^{\color{red}2n}}{(2n)!}}{2}dx$

Huh ?

Thank you...I had the correct way on my paper but I just mistyped it....and I looked off of my typing...so I just kept copying my mistake...so the way I have it now is the way it is ony my paper...so that isnt my mistake if there is one....is there any other mistakes?

**Moo** · April 17th 2008, 06:50 AM

Why should there be an error ?

Try factoring by x so that you have x^(4n)

x^(4n)=(x²)^(2n)

Try to put everything in that power, putting away constants that are not useful..
You got (2n)! on the denominator, so maybe it'll be something like cosine...

**angel.white** · April 17th 2008, 07:24 AM

I don't think that
$\int{cos(mx^2)^2dx} = \int cos^2(mx^2) ~dx$

I think it it should be
$\int{cos(mx^2)^2dx} = \int cos(m^2x^4) ~dx$

Because I've gotten demerited on a test before for assuming the former (though that was with the natural log function, not a trig function).

**Moo** · April 17th 2008, 07:29 AM

It wouldn't be very logical... I mean why would they have written (mx²)² instead of m²x^4 ?

**angel.white** · April 17th 2008, 07:36 AM

Originally Posted by Moo

It wouldn't be very logical... I mean why would they have written (mx²)² instead of m²x^4 ?

I argued it too ^_^ (didn't get the points back, though)

Best I can figure is that the function applies to the funciton directly to it's right, In this case, that means f(x) = x^2, and g(x) = mx^2 so this is cos f(g(x))

This may not be what the problem is, IDK, but since the OP doesn't include the answer, I figured it could be, since I've experienced similar frustration in the past.

**Mathstud28** · April 17th 2008, 11:49 AM

Originally Posted by angel.white

I don't think that
$\int{cos(mx^2)^2dx} = \int cos^2(mx^2) ~dx$

I think it it should be
$\int{cos(mx^2)^2dx} = \int cos(m^2x^4) ~dx$

Because I've gotten demerited on a test before for assuming the former (though that was with the natural log function, not a trig function).

You know what I think I am right...and it was $\int{cos^2(mx^2)}$ ....you want to know how? haha I made up the problem...it was just for fun

**angel.white** · April 17th 2008, 12:10 PM

Originally Posted by Mathstud28

You know what I think I am right...and it was $\int{cos^2(mx^2)}$ ....you want to know how? haha I made up the problem...it was just for fun

Fair enough.

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