Integral :)
\[\int\limits (x-1)\sin(\Pi x)\]
Is that pi?
taht is Pi :)... so if I make dv=Sin(Pi*x), can I take out Pi? so v=-Picos(x)?
Basically. The derivative of sin(ax) = -a(cos(ax)), after all. And then you just do the rest of integration by parts.
oh cool :) I'll work on it and maybe you can help me by checking?
Sure, no problem.
@greenglasses like this?
I think I found a mistake on that ;/
Yeah, I'm not sure where you got the 2pi.
it should be (Pi)^2
I'm getting (x-1)(-pi cos(pi.x)) -sin(pi.x) + c
the v=-Pi(Cos[Pi*x])
Yeah, the integral of cos(pi.x) would be (1/pi)(sin(pi.x)), not pi(sin(pi.x)). The two pis would cancel out.
hmm I see... so then in your second comment, did you mean derivative or integral of Sin(ax)=-aCox(ax)?
I got confused lol
Second comment I meant derivative.
It's the opposite when you're going the other way 'round.
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