Question

How do you evaluate the definite integral `xsin 2x dx` from 0 to `pi/2`?

Answer 1

I would first use Integration by Parts to solve the indefinite integral and then apply the Fundamental Theorem of Calculus:
Integration by Parts:
`intf(x)g(x)dx=F(x)g(x)-intF(x)g'(x)dx`
Where:
`F(x)=intf(x)dx`
`g'(x)=(dg(x))/dx`
Let us choose:
`f(x)=sin(2x)`
`g(x)=x`
Applying Integration by Parts you get:
`x*-cos(2x)/2-int-cos(2x)/2*1dx=`
`-xcos(2x)/2+sin(2x)/4`
Now the Fundamental Theorem of Calculus:
`int_a^bf(x)dx=F(b)-F(a)`
In your case:
`-xcos(2x)/2+sin(2x)/4|_0^(pi/2)` `=`
`=[-pi/2cos(pi)/2+sin(pi)/4]-0=`
`=pi/4`