integration by parts (1 Viewer)

kooltrainer · Jun 22, 2010

integrate (from 0-1) x*sin(n*x*Pi/2) dx

tell me if u get what i got:
{[ 4sin (nPi/2) ]/ (nPi)^2} - [2cos(nPi/2)]/(nPi)]

n is just an integer constant

Rezen · Jun 22, 2010

I got,

$\frac{4sin(\frac{n\pi}{2})}{n^2\pi^2}-\frac{2cos(\frac{n\pi}{2})}{n\pi}$

jet · Jun 22, 2010

Remember that $\cos \frac{n\pi}{2}$ is always 0 and $sin\frac{n\pi}{2}$ is always 1 or -1 depending on the value of n. (1 if n = 1, 5, 9, ... and -1 if n = 3, 7, 11, ...) so you can simplify your answer considerably.

MOP777 · Jun 22, 2010

he said that n was only a constant integer, that means that sin(npi/2) can be -1, 1 or 0, ie if n =2 its zero?

same with cos, isn't cos(npi/2) = 1 when n=0 or minus 1 when n=2?

shaon0 · Jun 22, 2010

MOP777 said:
he said that n was only a constant integer, that means that sin(npi/2) can be -1, 1 or 0, ie if n =2 its zero?

same with cos, isn't cos(npi/2) = 1 when n=0 or minus 1 when n=2?

Yeah.
For n even, cos(n*pi/2)=+-1 and sin(n*pi/2)=0.
For n odd, cos(n*pi/2)=0 and sin(n*pi/2)=+-1.

jet · Jun 22, 2010

shaon0 said:
Yeah.
For n even, cos(n*pi/2)=+-1 and sin(n*pi/2)=0.
For n odd, cos(n*pi/2)=0 and sin(n*pi/2)=+-1.

whoops forgot about the others lol

Trebla · Jun 22, 2010

$u=x\quad du=dx\\\\dv=\sin\dfrac{n\pi x}{2}\,dx\quad v = -\dfrac{2}{n\pi}\cos\dfrac{n\pi x}{2}\\\\\int_{0}^{1}x\sin\dfrac{n\pi x}{2}\,dx\\\\=-\left [\dfrac{2x}{n\pi}\cos\dfrac{n\pi x}{2} \right ]_0^1+\dfrac{2}{n\pi}\int_{0}^{1}\cos\dfrac{n\pi x}{2}\,dx\\\\= -\dfrac{2}{n\pi}\cos\dfrac{n\pi}{2}+\dfrac{4}{n^2\pi^2}\left [\sin\dfrac{n\pi x}{2} \right ]_0^1\\\\= \dfrac{4}{n^2\pi^2}\sin\dfrac{n\pi}{2}-\dfrac{2}{n\pi}\cos\dfrac{n\pi}{2}$

which is what Rezen got...

That's the most simplified you can get. Considering cases of different values of n just makes things unnecessarily more complicated.

integration by parts (1 Viewer)

kooltrainer

New Member

Rezen

Member

jet

Banned

MOP777

New Member

shaon0

...

jet

Banned

Trebla

Administrator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)