To evaluate the integral problem: %7D%7D%7D%7B%5Cleft.%7Bd%7D%7Bx%7D%5Cright.%7D "int_0^(pi/2) xsin(2x) dx")
,we may first solve for its indefinite integral. Indefinite integral are written in the form of %7D%7D%7D%7B%5Cleft.%7Bd%7D%7Bx%7D%5Cright.%7D%3D%7BF%7D%7B%5Cleft(%7Bx%7D%5Cright)%7D%2B%7BC%7D "int f(x) dx = F(x) +C")
where: %7D%7D%7D "f(x)")
as the integrand
%7D "F(x)")
as the anti-derivative function
as the arbitrary constant known as constant of integration
We follow a formula from basic integration table to determine the indefinite integral function ...
To evaluate the integral problem: ,we may first solve for its indefinite integral. Indefinite integral are written in the form of
where: as the integrand
as the anti-derivative function
as the arbitrary constant known as constant of integration
We follow a formula from basic integration table to determine the indefinite integral function . For the integrals with logarithm, the problem resembles the formula:
%7D%7D%7D%7B%5Cleft.%7Bd%7D%7Bx%7D%5Cright.%7D%3D-%5Cfrac%7B%7B%7Bx%7D%7B%5Ccos%7B%7B%5Cleft(%7Ba%7D%7Bx%7D%5Cright)%7D%7D%7D%7D%7D%7B%7Ba%7D%7D%2B%5Cfrac%7B%7B%5Csin%7B%7B%5Cleft(%7Ba%7D%7Bx%7D%5Cright)%7D%7D%7D%7D%7B%7B%7Ba%7D%7D%5E%7B%7B2%7D%7D%7D%2B%7BC%7D "int x sin(ax) dx= -(xcos(ax))/a+sin(ax)/a^2 +C")
.
By comparing %7D%7D%7D "x sin(ax)")
with%7D%7D%7D "xsin(2x)")
, we determine that 
.
Plug-in to the integral formula, we get:
%7D%7D%7D%7B%5Cleft.%7Bd%7D%7Bx%7D%5Cright.%7D%3D-%5Cfrac%7B%7B%7Bx%7D%7B%5Ccos%7B%7B%5Cleft(%7B%5Cleft(%7B2%7D%5Cright)%7D%7Bx%7D%5Cright)%7D%7D%7D%7D%7D%7B%7B%7B2%7D%7D%7D%2B%5Cfrac%7B%7B%5Csin%7B%7B%5Cleft(%7B%5Cleft(%7B2%7D%5Cright)%7D%7Bx%7D%5Cright)%7D%7D%7D%7D%7B%7B%7B%5Cleft(%7B2%7D%5Cright)%7D%7D%5E%7B%7B2%7D%7D%7D%7B%7B%5Cmid%7D_%7B%7B0%7D%7D%5E%7B%7B%5Cfrac%7B%5Cpi%7D%7B%7B2%7D%7D%7D%7D%7D "int_0^(pi/2) xsin(2x) dx=-(xcos((2)x))/(2)+sin((2)x)/(2)^2|_0^(pi/2)")
%7D%7D%7D%7D%7D%7B%7B2%7D%7D%2B%5Cfrac%7B%7B%5Csin%7B%7B%5Cleft(%7B2%7D%7Bx%7D%5Cright)%7D%7D%7D%7D%7B%7B4%7D%7D%7B%7B%5Cmid%7D_%7B%7B0%7D%7D%5E%7B%7B%5Cfrac%7B%5Cpi%7D%7B%7B2%7D%7D%7D%7D%7D "=-(xcos(2x))/2+sin(2x)/4|_0^(pi/2)")
After solving the indefinite integral from, we may apply definite integral formula:%7D%7B%7B%5Cmid%7D_%7B%7Ba%7D%7D%5E%7B%7Bb%7D%7D%7D%3D%7BF%7D%7B%5Cleft(%7Bb%7D%5Cright)%7D-%7BF%7D%7B%5Cleft(%7Ba%7D%5Cright)%7D "F(x)|_a^b = F(b) - F(a)")
.
%7D%7D%7D%7D%7D%7B%7B2%7D%7D%2B%5Cfrac%7B%7B%5Csin%7B%7B%5Cleft(%7B2%7D%7Bx%7D%5Cright)%7D%7D%7D%7D%7B%7B4%7D%7D%7B%7B%5Cmid%7D_%7B%7B0%7D%7D%5E%7B%7B%5Cfrac%7B%5Cpi%7D%7B%7B2%7D%7D%7D%7D%7D%3D%7B%5Cleft%5B-%5Cfrac%7B%7B%7B%5Cleft(%5Cfrac%7B%5Cpi%7D%7B%7B2%7D%7D%5Cright)%7D%5Ccdot%7B%5Ccos%7B%7B%5Cleft(%7B2%7D%5Ccdot%7B%5Cleft(%5Cfrac%7B%5Cpi%7D%7B%7B2%7D%7D%5Cright)%7D%5Cright)%7D%7D%7D%7D%7D%7B%7B2%7D%7D%2B%5Cfrac%7B%7B%5Csin%7B%7B%5Cleft(%7B2%7D%5Ccdot%7B%5Cleft(%5Cfrac%7B%5Cpi%7D%7B%7B2%7D%7D%5Cright)%7D%5Cright)%7D%7D%7D%7D%7B%7B4%7D%7D%5Cright%5D%7D-%7B%5Cleft%5B-%5Cfrac%7B%7B%7B0%7D%5Ccdot%7B%5Ccos%7B%7B%5Cleft(%7B2%7D%5Ccdot%7B0%7D%5Cright)%7D%7D%7D%7D%7D%7B%7B2%7D%7D%2B%5Cfrac%7B%7B%5Csin%7B%7B%5Cleft(%7B2%7D%5Ccdot%7B0%7D%5Cright)%7D%7D%7D%7D%7B%7B4%7D%7D%5Cright%5D%7D "-(xcos(2x))/2+sin(2x)/4|_0^(pi/2) =[-((pi/2) *cos(2*(pi/2)))/2+sin(2*(pi/2))/4]-[-(0*cos(2*0))/2+sin(2*0)/4 ]")
%7D%5Ccdot%7B%5Ccos%7B%7B%5Cleft(%5Cpi%5Cright)%7D%7D%7D%7D%7D%7B%7B2%7D%7D%2B%5Cfrac%7B%7B%5Csin%7B%7B%5Cleft(%5Cpi%5Cright)%7D%7D%7D%7D%7B%7B4%7D%7D%5Cright%5D%7D-%7B%5Cleft%5B-%5Cfrac%7B%7B%7B0%7D%5Ccdot%7B%5Ccos%7B%7B%5Cleft(%7B0%7D%5Cright)%7D%7D%7D%7D%7D%7B%7B2%7D%7D%2B%5Cfrac%7B%7B%5Csin%7B%7B%5Cleft(%7B0%7D%5Cright)%7D%7D%7D%7D%7B%7B4%7D%7D%5Cright%5D%7D "=[-((pi/2) *cos(pi ))/2+sin(pi) /4]-[-(0*cos(0))/2+sin(0)/4 ]")
%7D%7D%7D%7B%7B4%7D%7D%2B%5Cfrac%7B%7B0%7D%7D%7B%7B4%7D%7D%5Cright%5D%7D-%7B%5Cleft%5B-%5Cfrac%7B%7B%7B0%7D%5Ccdot%7B1%7D%7D%7D%7B%7B2%7D%7D%2B%5Cfrac%7B%7B%7B0%7D%7D%7D%7B%7B4%7D%7D%5Cright%5D%7D "=[-(pi*(-1))/4+0 /4]-[-(0*1)/2+(0)/4 ]")
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