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The exact question is to take the Integral between 0 and pi of sine^5 x dx I know that it needs to be solved using u-substitution.

2006-09-27 16:16:43 · 3 answers · asked by garrettmyers1989 1 in Science & Mathematics Mathematics

3 answers

(sin x)^5 dx = (sin x)^2 (sin x)^2 (sin x) dx
= (1 - (cos x)^2)(1 - (cos x)^2) (sin x) dx
Let u = cos x, then du = -sin x dx
Now you have
-(1-u^2)(1-u^2) du
Expand and integrate
Also, notice that sin(0) = sin(pi) = 0
You will need to double the integral and only integrate from 0 to pi/2

2006-09-27 16:28:43 · answer #1 · answered by MsMath 7 · 1 0

sin^5 x = (sin^2x)^2 . sin x
=(1-cos^2x)^2 .sin x

put cos x = u
- sin x dx = du
so integral
limit of u is from 1 to -1 (lowe limit 1 higer -1)
= int(-(i-u^2)^2) du

= int(-1+2u^2-u^4) from 1 to -1

-u + u^3(2/3) - u^5/5

at 1 it is -1 + 2/3 - 1/5 = 2/3 - 6/5 = -8/15
at -1 = -1 -2/3 +1/5 = 1/5-5/3 = -25/15

so the value = (-25-8)/15 or -33/15

2006-09-27 17:23:04 · answer #2 · answered by Mein Hoon Na 7 · 0 0

16/15

2006-09-27 16:24:44 · answer #3 · answered by عبد الله (ドラゴン) 5 · 0 2

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