maths definite integral

1 ∫(x=0 to π) xf(sinx) dx

Let y = π - x, x = 0, y = π. x = π, y = 0

= - ∫(y=π to 0) (π - y) f(siny) dy

= ∫(y= 0 to π) (π - y) f(siny) dy

So,

∫(x=0 to π) xf(sinx) dx= π/2 ∫(x=0 to π) f(sinx) dx.

2 ∫(x=0 to a) f(x)g(x) dx=K/2 ∫(x=0 to a)f(x) dx,

f(x)=f(a-x) and g(x)+g(a-x)=K,where K is a constant.

Let a = π , f(x) = (sinx)^2, g(x) = x(cosx)^4. K = π

So, ∫(x=0 to π) xsin^2xcos^4x dx

= π/2





2011-04-30 16:54:34 補充：
f(x) = (sinx)^2(cosx)^4 ,g(x) = x

∫(x=0 to a) f(x)g(x) dx

= π/2 ∫(x=0 to π) (sinx)^2(cosx)^4 dx

= π/16 ∫(x=0 to π) (sin2x)^2(1 + cos2x) dx

= π/32 ∫(y=0 to 2π) (siny)^2(1 + cosy) dy

= π/32 ∫(y=0 to 2π) (1 - sin2y)/2 + (siny)^2(cosy) dy

= (π/32)[y/2 - cos2y/4 + (siny)^3/3] |[0,2π]

= π^2/32