Fairly hard Integration Question!!! (1 Viewer)

EvoRevolution · Mar 21, 2009

Ok will when i was trying this i got nearly 2 pages of working out and was nearly right i think i made a couple of mistakes and didnt want to recheck.

Drongoski · Mar 21, 2009

EvoRevolution said:
Ok will when i was trying this i got nearly 2 pages of working out and was nearly right i think i made a couple of mistakes and didnt want to recheck.

Does the integral not have limits of integration ??

youngminii · Mar 21, 2009

It's a fairly simple recurrence formula question. Just use integration by parts..

Drongoski · Mar 21, 2009

kwabon said:
could someone please show me the working out

I will! I've just finished it. But I'm very slow @ LaTeX . . so give me some time.

kwabon · Mar 21, 2009

nuts of a question this .... i cant get it into the form

Drongoski · Mar 21, 2009

$I_{n}\ \ =\ \ \int x^{-n}sinxdx\\ \\ \ =\ \ \ \frac{1}{-n + 1}\int sinx\ d(x^{-n+1})\\ \\ =\ \ \ \frac{1}{(1 - n)}\ \left [x^{-n+1}sinx\ \ -\ \ \int x^{-n+1}d\ sinx \right ]\\ \\=\ \ \ \frac{1}{(1-n)}\ \ \left [x^{-n+1}sinx\ \ - \int x^{-x+1}cosx\ dx \right ]\\ \\ =\ \ \ \frac{1}{(1-n)}\ \ \ \left [x^{-n+1}sinx\ \ -\ \ \frac{1}{-n+2}\ \int cosx\ dx^{-n+2} \right ]\\ \\ =\ \ \ \frac{1}{(1-n)}\ \ \ \left [x^{-n+1}sinx\ \ +\ \ \frac{1}{(n-2)} \right\ \ \left \{x^{-n+2}\ cosx\ \ -\ \ \int x^{-n+2}\ d\ cosx \right \}]\\ \\=\ \ \ \frac{1}{(1-n)}\ \ \ \left [x^{-n+1}sinx\ \ + \ \ \frac{1}{(n-2)} \right\ \left \{x^{-n+2}cosx\ \ +\ \ \int x^{-n+2}sinx\ dx \right \} ]\\ \\ =\ \ \ \frac{1}{(1-n)}\ \ \left [x^{-n+1}sinx\ \ +\ \ \frac{1}{(n-2)}\left \{x^{-n+2}cosx\ \ +\ \ I_{n-2} \right \}} \right ]\\ \\ \\ After\ some\ fiddling\ you\ end\ up\ with\ \\ \\ (1-n)(n-2) I_{n}\ \ \ =\ \ \ x^{-n+1}\ \left [xcosx\ \ +\ (n-2)sinx \right ]\ \ + \ \ I_{n-2}\\ \\QED\\ \\$

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Timothy.Siu · Mar 21, 2009

i just did this easily...

not hard at all.
u can just do it by parts twice fairly quickly, just make sure u integrate the x^-n but u can tell from the question anyway because u have to get I^n-2

EvoRevolution · Mar 21, 2009

can some1 post the full solution plz.
cause i think it doesnt really solve so easily.

Trebla · Mar 21, 2009

Didn't Drongoski already did that?

EvoRevolution · Mar 21, 2009

he solved half

EvoRevolution · Mar 21, 2009

umm really comfused on how u went from the original to the 2nd step
plz explain

Timothy.Siu · Mar 21, 2009

EvoRevolution said:
umm really comfused on how u went from the original to the 2nd step
plz explain

he did it by parts

Trebla · Mar 21, 2009

If you can't understand Drongoski's working, then try looking at this:
$I_n=\displaystyle\int x^{-n}\sin x\,dx\\u = \sin x, du = \cos x\,dx\\dv=x^{-n}\,dx, v = \dfrac{x^{-n+1}}{-n+1}\\\\I_n=\dfrac{x^{-n+1}}{1-n}\sin x - \dfrac{1}{1-n}\displaystyle\int x^{-n+1}\cos x\,dx\\u = \cos x, du = -\sin x\,dx\\dv=x^{-n+1}\,dx, v = \dfrac{x^{-n+2}}{-n+2}\\\\I_n=\dfrac{x^{-n+1}}{1-n}\sin x - \dfrac{1}{1-n}\left [\dfrac{x^{-n+2}\cos x}{-n+2}+\dfrac{1}{-n+2}\displaystyle\int x^{-(n-2)}\sin x\,dx\right]\\\\(1-n)I_n=x^{-n+1}\sin x - \dfrac{x^{-n+2}\cos x}{-n+2}-\dfrac{1}{-n+2}I_{n-2}\\\\(1-n)I_n=x^{-n+1}\sin x + \dfrac{x^{-n+2}\cos x}{n-2}+\dfrac{1}{n-2}I_{n-2}\\\\(1-n)(n-2)I_n=(n-2)x^{-n+1}\sin x + x^{-n+2}\cos x+I_{n-2}\\\\(1-n)(n-2)I_n=x^{-n+1}\left [x\cos x + (n-2)\sin x\right ]+I_{n-2}\\$

Drongoski · Mar 21, 2009

EvoRevolution · Mar 21, 2009

ok kool thanks guys makes sense

EvoRevolution · Mar 21, 2009

i see where i went wrong i just make some really stupid mistakes so ended up goin really bad for me.

first attempt i let u = x^-n

that went horrible i ended up with 2 pages for workin out and no answer

Fairly hard Integration Question!!! (1 Viewer)

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