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