The answer is: 1/16(pi^2+4)116(π2+4)
The first thing to do is to lower the degree of the function sinus with the Half-angle formula:
sinx=sqrt((1-cos2x)/2sinx=√1−cos2x2.
So the integral becomes:
int_0^(pi/2)x(1-cos2x)/2dx=1/2int_0^(pi/2)(x-xcos2x)dx=1/2[int_0^(pi/2)xdx-int_0^(pi/2)xcos2xdx]∫π20x1−cos2x2dx=12∫π20(x−xcos2x)dx=12[∫π20xdx−∫π20xcos2xdx].
To do the first integral it is sufficient to remember the law:
intx^ndx=x^(n+1)/(n+1)+c∫xndx=xn+1n+1+c,
so:
int_0^(pi/2)xdx=[x^2/2]_0^(pi/2)=[(pi^2/4)/2-0/2]=pi^2/8∫π20xdx=[x22]π20=⎡⎣π242−02⎤⎦=π28
The second integral can be done using the theorem of the integration by parts, that says:
intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx
We can assume that f(x)=x and g'(x)dx=cos2xdx.
To find g(x), it is sufficient to remember this law:
g(x)=intcos(h(x))h'(x)dx=sin(h(x))+c,
where h(x)=2x, and h'(x)=2,
So:
g(x)=intcos2xdx=1/2intcos2x.2dx=1/2sin2x.
and f'(x)=1
So:
int_0^(pi/2)xcos2xdx=[x*1/2sin2x]_0^(pi/2)-int_0^(pi/2)1*1/2sin2xdx=
[pi/2*1/2sin2(pi/2)-0]-1/2[-(cos2x)/2]_0^(pi/2)=[pi/4sinpi+1/2(cos2(pi/2))/2-1/2cos0/2]=[0-1/4-1/4]=-1/2.
I remember that intsin2xdx=-(cos2x)/2, done like before with intcos2xdx=(sin2x)/2
Final count:
1/2(pi^2/8+1/2)=1/16(pi^2+4)
