.
int_0^1xsqrt(1-x^4)dx
u=x^2
x=sqrtu
u^2=x^4
du=2xdx
dx=(du)/(2x)=(du)/(2sqrtu)
sqrt(1-x^4)=sqrt(1-u^2)
Now, we substitute the above into the problem integral to turn it into an integral in terms of u:
int(sqrtu)(sqrt(1-u^2))((du)/(2sqrtu))=
1/2intsqrt(1-u^2)du
Now we can use trigonometric substitution to solve this:
u=sintheta
du=costhetad(theta)
1/2intsqrt(1-sin^2theta)costhetad(theta)
1/2intsqrt(cos^2theta)costhetad(theta)
1/2intcosthetacosthetad(theta)=1/2intcos^2thetad(theta)
cos^2theta=(1+cos(2theta))/2
1/4int(1+cos(2theta))d(theta)=1/4intd(theta)+1/4intcos(2theta)d(theta)
1/4theta+1/8sin(2theta)
Now, we can substitute back for u:
u=sintheta, then theta=arcsinu
costheta=sqrt(1-six^2theta)=sqrt(1-u^2)
sin(2theta)=2sinthetacostheta=2usqrt(1-u^2)
1/4theta+1/8sin(2theta)=(arcsinu+usqrt(1-u^2))/4
Now, we can substitute back for x:
int_0^1xsqrt(1-x^4)dx=(arcsinx^2+x^2sqrt(1-x^4))/4
We will evaluate this from 0 to 1:
(arcsin1^2+1^2sqrt(1-1^4))/4-(arcsin0^2+0^2sqrt(1-0^4))/4=
(pi/2+0)/4-(0+0)/4=pi/8