Using integration by parts, show that integration x squared cos 2x dx with limits pi and 0 is equal to pi/2?
Answers:
email me, calcu_lust@yahoo.com, i'll write out a nice solution.
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Int(x^2.cos2xdx)
u = x^2 => du = 2x.dx
dv = cos2x.dx => v = (1/2).sin2x
=>Int(u.dv) = uv - Int(v.du) = 2.x^2.sin2x {0 -> pi} - Int(2x.(1/2).sin2x.dx) = 0 - Int(x.sin2x.dx)
Now calculate Int(x.sin2x.dx) {0 -> pi}
Call u = x => du = dx
Call dv = sin2x.dx => v = -cos2x/2
Int(x.sin2x.dx) = uv {0 -> pi} - Int(v.du) = -x.cos2x/2 {0 -> pi} - Int(-cos2x/2.dx) = -(pi.cos2pi)/2 + 1/4 sin2x {0->pi} = -pi/2
=> Int(x^2.cos2x.dx) {0 ->pi} = - Int(x.sin2x.dx) {0 -> pi} = pi/2
first,work out indefinite integral
let G=int(x^2*cosxdx)
using integration by parts,
u= X^2 and dv/dx =cos2x
>>>> dv=cos2x*dx
so that,v=int(cos2xdx)
= 1/2 sin2x
du/dx=2x
using the formula for integration
by parts
int(udv)=uv-int(vdu)
G=1/2x^2*sin2x
-int(1/2sin2x*2x*dx)
=1/2x^2*sin2x
-int(x*sin2x*dx){1}
find int(x*sin2x*dx)
U=x, and dV/dx=sin2x
dU/dx=1, V=int(sin2xdx)
= -1/2cos2x
int(x*sin2x*dx)
= -1/2x*cos2x-int(1/2cos2x*1dx)
= -1/2x*cos2x+1/4sin2x
substitute back into{1}
G=1/2x^2*sin2x
-[ -1/2x*cos2x+1/4sin2x]
=1/2x^2*sin2x+1/2x*cos2x
-1/4sin2x
=1/2x*cos2x+1/4(2x^2-1)sin2x+C
we are now setting limits,so
C vanishes
therefore,G for x between pi and 0
= [1/2pi*(1)+0 ]-0
=1/2pi as required
i hope that this helps
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