How do you find the integral intx*sqrt(1-x^4)dx ?

1 Answer

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