[Sig]

Targget:

......Pi/2
Kn={ (sinx)^n dx
......0

{- integral sign The integral is definate, has Pi/2 on the top, and 0 on the botom.
Pi- pie radians= 180
Kn- K based n
P.S. Sorry for the crapy notation

Objective: Obtain a reduction formula for Kn with n>1, and hence evaluate K4 and K5.

I know it will be hard to solve this on a comp, but any help is appriciated, also give any revision notes for this topic, if u can.

Thanks for the effort

[notnek]

See if you can finish what I started.

[SunGod87]

Alternatively split it into sin^(n-1)x sinx.

Spoiler:

Then integrate by parts

[kat2pult]

Haha I didn't know whether to read that as 'hardcore' integration, or core intregration which is hard!

[-G-a-v-]

Originally Posted by kat2pult

Haha I didn't know whether to read that as 'hardcore' integration, or core intregration which is hard!

Reduction formulae is a further maths topic, so probably the former

[Sig]

See if you can finish what I started.

Originally Posted by SunGod87

lternatively split it into sin^(n-1)x sinx.

Ok guys, with u'r help I did this...
.................................................. .......pi/2
{sin^(n-2)x cos^2x dx= -[cosx sin^(n-1)x] + {sin^(n-2)x cos^2x dx=
.................................................. .......0
(cos^2x=1-sin^2x), so
.........................pi/2..............pi/2...........................pi/2
=-[cosx sin^(n-1)x] + (n-1) {sin^(n-2)x dx - (n-1) {sin^2x dx
.........................0..................0..... .........................0

Kn=(n-1) Kn-2 - (n-1) K2

Where Kn-2 is K to the base (n-2), K2 is K based 2.

Is that correct though?
0
Note: { - Integral, The integral is definate, has Pi/2 on the top, and 0 on the botom. so [cosx sin^(n-1)x] has i/2 on the top, and 0 on the botom.

[DFranklin]

I don't think so, but what you've posted is near illegible, and has no explanation, which doesn't help.

I think SunGod's suggestion is better than Notnek's, for what it's worth.

[notnek]

I agree with DFranklin - follow SunGod's method .