reduction formula (1 Viewer)

[ezzy85]

how would you do this:
I = cosⁿx e^x dx

im having a total mental blank.
thanks

eidt:the target is: In = (n(n-1)/[n² + 1]) * I_n-2 and the limits are -pi/2 to pi/2

 

[ngai]

normally i don't bother with this...so much to type...

In = cos^n x e^x dx
= [e^x cos^n x] - int{ e^x n.cos^(n-1)x . sinx } dx (int by parts)
= 0 - nint{ e^x cos^(n-1)x . sinx } dx
= 0 - n[e^x cos^(n-1)x . sinx] + n.int{e^x[cos^(n-1)x.cosx - sinx.(n-1).cos^(n-2)x .sinx} dx
= 0 - 0 + n.int{e^x.cos^n x} dx - n(n-1)int{cos^(n-2)x (1-cos^2 x)} dx
= 0 - 0 + nIn - n(n-1)int{cos^(n-2)x - cos^n x)} dx
= 0 - 0 + nIn - n(n-1)I[n-2] + n(n-1)In
Therefore:
(1 - n - n^2 + n)In = -n(n-1)I[n-2]
(n-1)(n+1)In = n(n-1)In-2
(n+1)In = nIn-2
In = [n/(n+1)]In-2

hmm, doesn't match ur answer, so maybe i made error in all tha ttyping
u check it, but the basic method is there

 

[Archman]

ah the typo was in line 2

In = cos^n x e^x dx
= [e^x cos^n x] - int{ e^x n.cos^(n-1)x . sinx } dx (no it should be -sinx)
so it should be
= [e^x cos^n x] + int{ e^x n.cos^(n-1)x . sinx } dx
after pretty much the same working out...
= 0 + 0 - nIn + n(n-1)I[n-2] - n(n-1)In
Therefore:
(1 + n + n^2 - n)In = n(n-1)I[n-2]
(1 + n^2)In = n(n-1)In-2
as desired

 

[ezzy85]

i see.. thnaks guys.