integral (1-cos x)sin x dx from 0 to pi
-
Use a substitution:
u = 1 − cosx
du = sinx dx
When x = 0, u = 1 − cos(0) = 1 − 1 = 0
When x = π, u = 1 − cos(π) = 1 − (−1) = 2
∫₀ᵖⁱ (1 − cosx) sinx dx
= ∫₀² u du
= 1/2 u² |₀²
= 1/2 (4−0)
= 2
u = 1 − cosx
du = sinx dx
When x = 0, u = 1 − cos(0) = 1 − 1 = 0
When x = π, u = 1 − cos(π) = 1 − (−1) = 2
∫₀ᵖⁱ (1 − cosx) sinx dx
= ∫₀² u du
= 1/2 u² |₀²
= 1/2 (4−0)
= 2
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integral (1-cosx)sinx dx =integral sinx dx - integral cosx*sinx*dx
integral sinx = -cosx
cosx*sinx = (1/2)*2*sinx*cosx= (1/2)sin(2x)
integral (1/2)sin(2x)dx = (1/2)integral sin(2x)*dx
(1/2)*(1/2)*integral sin(2x)*2*dx
(1/4)(-cos(2x)
integral = [ -cosx - (1/4)cos(2x) ] from 0 to pi =
- [cos(pi) + (1/4)cos(2pi) -cos(0) - (1/4)cos(2*0) ]
-[ -1 +(1/4)*1 - 1 -(1/4)*1 ] = 2
integral sinx = -cosx
cosx*sinx = (1/2)*2*sinx*cosx= (1/2)sin(2x)
integral (1/2)sin(2x)dx = (1/2)integral sin(2x)*dx
(1/2)*(1/2)*integral sin(2x)*2*dx
(1/4)(-cos(2x)
integral = [ -cosx - (1/4)cos(2x) ] from 0 to pi =
- [cos(pi) + (1/4)cos(2pi) -cos(0) - (1/4)cos(2*0) ]
-[ -1 +(1/4)*1 - 1 -(1/4)*1 ] = 2