Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 42

To evaluate the integral problem: int_0^(pi/2) xsin(2x) dx ,we may first solve for its indefinite integral. Indefinite integral are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function
C as the arbitrary constant known as constant of integration
We follow a formula from basic integration table to determine the indefinite integral function F(x) . For the integrals with logarithm, the problem resembles the formula:
int x sin(ax) dx= -(xcos(ax))/a+sin(ax)/a^2 +C .
By comparing x sin(ax) with xsin(2x) , we determine that a= 2 .
Plug-in a=2 to the integral formula, we get:
int_0^(pi/2) xsin(2x) dx=-(xcos((2)x))/(2)+sin((2)x)/(2)^2|_0^(pi/2)
=-(xcos(2x))/2+sin(2x)/4|_0^(pi/2)
After solving the indefinite integral from, we may apply definite integral formula: F(x)|_a^b = F(b) - F(a) .
-(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 ]
=[-((pi/2) *cos(pi ))/2+sin(pi) /4]-[-(0*cos(0))/2+sin(0)/4 ]
=[-(pi*(-1))/4+0 /4]-[-(0*1)/2+(0)/4 ]
=[pi/4+0]-[0+0]
= [pi/4] - [0]
=pi/4