Question about reduction formula in integration !?

I_1 = integral(sin[x])dx from 0 to pi/2 = -1 * cosx from 0 to pi/2 = -1*(cos[pi/2] - cos[0]) = -1*(0-1) = -1*(-1) = 1

further on we'll use sin^2[x] = (1-cos[2x])/2 = 1/2 - (cos[2x])/2

I_2 = integral(sin^2[x])dx from 0 to pi/2 = integral(1/2 - (cos[2x])/2)dx from 0 to pi/2 =

. . .= integral(1/2)dx from 0 to pi/2 - integral{(cos[2x])/2}dx from 0 to pi/2 =

. . .= x/2 from 0 to pi/2 - (1/2)*integral{cos[2x]}dx from 0 to pi/2 =

. . .= x/2 from 0 to pi/2 - (1/2)*(1/2)integral{sin'[2x]}dx from 0 to pi/2 =

. . .= (1/2)*(pi/2 - 0) - (1/4)*sin[2x] from 0 to pi/2 =

. . .= pi/4 - (1/4)*{sin[2*(pi/2)] - sin[2*0] } =

. . .= pi/4 - (1/4)*{sin[pi] - sin[0] } =

. . .= pi/4 - (1/4)*(0-0) = pi/4 - (1/4)*0 = pi/4

I did this because there is a general formula that you can easily obtain through integration by parts. You write sin^n[x] = sin^(n-1)[x]*sin[x] = sin^(n-1)[x]*(-cos[x])' and from there it is easy. But it's a lot to write!

You get this recurring formula:

I_n = -{sin^(n-1)[x]*cos[x]} / n + [(n-1)/n] * I_(n-2) all from 0 to pi/2

I_n = -{sin^(n-1)[pi/2]*cos[pi/2] - sin^(n-1)[0]*cos[0] } / n + [(n-1)/n] * I_(n-2)

I_n = -{1*0 - 0*1 } / n + [(n-1)/n] * I_(n-2) from 0 to pi/2

I_n = -{0 - 0} / n + [(n-1)/n] * I_(n-2) from 0 to pi/2

I_n = 0 / n + [(n-1)/n] * I_(n-2) from 0 to pi/2

I_n = 0 + [(n-1)/n] * I_(n-2) from 0 to pi/2

I_n = [(n-1)/n] * I_(n-2) from 0 to pi/2

I_3 = [(3-1)/3] * I_1 = 2/3 * 1 = 2/3

I_4 = [(4-1)/4] * I_2 = 3/4 * pi/4 = 3pi/16

I_5 = ---> I don't care...

I_6 = [(6-1)/6] * I_4 = 5/6 * 3pi/16 = 15pi/96

Hope you understand!

Best regards!

PS: if you are lazy try this site: http://www.solvemymath.com/online_math_calculator/...

Just type [sin(x)]^6 and it will calculate the indefinite integral. After that just replace x with pi/2 then x with 0 and subtract the two results. you'll get exactly what I got. ;)

You need to use integration by parts. The fist time write sin^n as sin(x)*sin(n-1). Then you will end up with an integral a bit like: (n-1)*cos^2*sin^n-1 you will need to then use integration by parts agian on this. Then it becomes easy and you can find an expression for Isubn quite easily. It is alot of tedious algebra though.

Your question is incomplete unless you mention whether n is even or odd. I suppose, n is even.

∫ sin^n(x) dx from 0 to pi/2

= {pi/2}{(n - 1)(n - 3)(n - 5)....(5)(3)(1)}/{(n)(n - 2)(n - 4)....(6)(4)(2)}.