.
#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#
