Answer, #=1/4(sin^(-1)(x^2)+x^2sqrt(1-x^4))+c#
Explanation :
#intx*sqrt(1-x^4)dx = intx*sqrt(1-(x^2)^2)dx#
Using Trigonometric Substitution
let's #x^2=sint, => 2xdx=costdt#. Here I am using #sint#, it can also be done by considering #x^2=cost#.
Now plugging in integral,
#=int1/2sqrt(1-x^4)(2xdx)#
#=int1/2*cost*sqrt(1-sin^2t)dt#
Using the identity, #sin^2t+cos^2t=1#
we get,
#=int1/2*cos^2tdt#,
Using another important identity, #cos2t=cos^2t-sin^2t#
#=>cos2t=2cos^2t-1#
#=>cos^2t=(1+cos2t)/2#
#=int1/2*1/2*(1+cos2t)dt = 1/4int1dt+1/4intcos2tdt#
Using Trigonometric functions,
#=1/4*t+1/4(sin2t)/2+c#, where c is constant
#=1/4*(t+(sin2t)/2)+c#
Use the identity #sin2t=2sintcost#:
#=1/4(t+(2sintcost)/2)+c#
#=1/4(t+sintcost)+c#
#=1/4(t+sint sqrt(1-sin^2t))+c#
And substituting #sint=x^2# back in:
#=1/4(sin^(-1)(x^2)+x^2sqrt(1-x^4))+c#
