[tex]\int x^2sin\pixdx[\math]

why doesnt this latex thing work for me????

i dont know how to integrate x^2sin(pi)xdx

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- Jan 29th 2007, 01:58 PMjephIntegration by parts
[tex]\int x^2sin\pixdx[\math]

why doesnt this latex thing work for me????

i dont know how to integrate x^2sin(pi)xdx - Jan 29th 2007, 03:27 PMtopsquark
Your "slash" for the closing bracket is in the wrong direction:

$\displaystyle \int x^2 sin \pi xdx$

Using spaces would be a good idea, too.

$\displaystyle \int p dq = pq - \int q dp$

So

$\displaystyle \int x^2 sin(\pi x)dx$

Pick

$\displaystyle p = x^2$ ==> $\displaystyle dp = 2x dx$

$\displaystyle dq = sin(\pi x) dx$ ==> $\displaystyle q = -\frac{1}{\pi}cos(\pi x) $

Then

$\displaystyle \int x^2 sin(\pi x)dx = -\frac{1}{\pi} x^2 cos(\pi x) - \int (2x) \cdot -\frac{1}{\pi}cos(\pi x) dx$$\displaystyle = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2}{\pi} \int x cos(\pi x) dx$

We can use the same process on the remaining integral:

$\displaystyle \int x cos(\pi x) dx = \frac{x}{\pi}sin(\pi x) - \int \frac{1}{\pi}sin(\pi x) dx$$\displaystyle = \frac{x}{\pi}sin(\pi x) + \frac{1}{\pi ^2}cos(\pi x)$

Putting this result into our original integral:

$\displaystyle \int x^2 sin(\pi x)dx = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2}{\pi} \left ( \frac{x}{\pi}sin(\pi x) + \frac{1}{\pi ^2}cos(\pi x) \right )$

$\displaystyle = -\frac{1}{\pi} x^2 cos(\pi x) + \frac{2x}{\pi ^2}sin(\pi x) + \frac{2}{\pi ^3}cos(\pi x)$

-Dan - Jan 29th 2007, 05:20 PMjeph
- Jan 29th 2007, 07:03 PMtopsquark
That isn't quite what I did, you are off by a negative.

In detail:

$\displaystyle \int \frac{1}{\pi} sin(\pi x) dx = \frac{1}{\pi} \int sin(\pi x) dx$

Let $\displaystyle y = \pi x$ ==> $\displaystyle dy = \pi dx$

So

$\displaystyle \int \frac{1}{\pi} sin(\pi x) dx = \frac{1}{\pi} \int \frac{dy}{\pi}sin(y) = \frac{1}{\pi ^2}(-cos(y)) = -\frac{1}{\pi ^2}cos(\pi x)$

-Dan - Jan 29th 2007, 07:15 PMThePerfectHacker
A useful rule to remember....

Let $\displaystyle F(x)$ be an antiderivative of $\displaystyle f(x)$.

Then, ($\displaystyle a\not = 0$),

$\displaystyle \int f(ax+b)dx=\frac{1}{a}F(ax+b)+C$

Meaning whenever you have a linear function (of the form ax+b) you just take the coefficient with the "x" (the slope of line) and invert it.

For example,

$\displaystyle \int (2x+1)^4 dx$

Since this is a linear expression inside,

$\displaystyle 2x+1$.

You can use the rule I told you.

Just find the anti-derivative the regular way,

$\displaystyle \int (2x+1)^4 dx = \frac{(2x+1)^5}{5}+C$

But of course this answer is wrong, you need to invert the slope.

Thus,

$\displaystyle \frac{1}{2} \cdot \frac{(2x+1)^5}{5}+C$

Now let us find,

$\displaystyle \int \sin (\pi x) dx$

Note $\displaystyle \pi x=\pi x+0$ is a linear function in the special form we had above because $\displaystyle \pi$ is just a number in front.

Thus, to find the integral you just find the integral of sine which is -cosine and then invert the slope.

Thus,

$\displaystyle \int \sin \pi x dx = -\frac{1}{\pi} \cos x+C$.

Learn, this rule, it saves much time.