[tex]\int x^2sin\pixdx[\math]
why doesnt this latex thing work for me????
i dont know how to integrate x^2sin(pi)xdx
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
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
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.