Hard Integration q (1 Viewer)

Sirius Black

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prove that if

Let U<sub>n</sub>=<MATH>&int; xcos<sup>n</sup>x dx</MATH> from o to pi/2
then U<sub>n</sub>=(n-1)/n U<sub>n-2</sub>-1/n<sup>2</sup>
 

FinalFantasy

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Let Un=∫ xcos^n x dx from o to pi/2
Un=∫ xcos^n x dx=int.xcos^(n-1) x cos x dx
let u=cos^(n-1) x and dv\dx=xcosx
du\dx=(n-1)cos^(n-2)x(-sin x) and v=int. xcosx
du\dx=-(n-1)sinxcos^(n-2)x and v=xsinx+cosx

Un=[(xsinx+cosx)cos^(n-1)x] 0 to pi\2 +(n-1)int. (xsinx+cosx)sinxcos^(n-2)x dx
=-1+(n-1)int. xsin²xcos^(n-2)x dx+(n-1)int. sinxcos^(n-1)x dx
=-1+(n-1)int. xcos^(n-2)x dx-(n-1)int. xcos^n x dx+(n-1)int. sinxcos^(n-1) x dx
=-1+(n-1)U_(n-2)-(n-1)Un+(n-1)int. sinxcos^(n-1) x dx
Un(1+(n-1))=-1+(n-1)U_(n-2)+(n-1)int. sinxcos^(n-1) x dx
Un(n)=-1+(n-1)U_(n-2)+(n-1)int. sinxcos^(n-1) x dx
consider I=int. sinxcos^(n-1) x dx
let u=cos^(n-1)x dx and dv\dx=sinx
du\dx=(n-1)cos^(n-2)x(-sinx)=-sinx(n-1)cos^(n-2)x and v=-cosx
I=[-cos^n x] 0 to pi\2 -(n-1)int. sinxcos^(n-1) x dx
I(1+(n-1) )=1
I=1\n

back to:
Un(n)=-1+(n-1)U_(n-2)+(n-1)int. sinxcos^(n-1) x dx
Un(n)=-1+(n-1)U_(n-2)+(n-1)(1\n)
Un(n)=-1+(n-1)U_(n-2)+1-1\n
Un(n)=(n-1)U_(n-2)-1\n
Un=(n-1)\n U_(n-2)-1\n²

a bit messy, might of made error some where but looks rite lol
 

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