Integrals using U substitution?

Question

x/ Square root (1-x^4) dx

I let u = Square root (1-x^4)

du = 1/2(1-x^4)^ -1/2 * 4x^3 dx

How would i simplify for du?

Answer 1

substitute

u=x^2

du=2xdx or x dx=du/2

sqrt (1-x^4)=sqrt(1-u^2)

the expression is now

integration of (1/2)*(1/sqrt(1-u^2)*(du)

now put u=sink or k=sin inverse u

then du=cosk dk

1-u^2=1-sin^2k=cos^2k

sqrt(1-u^2)=sqrtcos^2k=cosk

the expression is now

integration of (1/2)(1/cosk)(coskdk)

=integration of (1/2)dk

=(1/2)k+ constant

=(1/2) (sin^(-1)u+ c

=(1/2)(sin^(-1)x^2+c ans

Answer 2

∫x dx / √(1 - x^4)

notice that x^4 = (x²)²

∫x dx / √[1 - (x²)²]

now let u = x²

du = 2x dx

x dx = (1/2) du

substitute:

∫(1/2)du / √(1-u²)

it's straight forward from here. The integral of ∫du / √(1-u²) = arcsin(u)

so (1/2) arcsin(u) + C

(1/2) arcsin(x²) + C

Answer 3

Better to replace x^2 by u. Then

du=2xdx=>xdx=0.5du

Your integral then turns to

0.5*integral(du/sqrt(1-u^2))

Further put u=sin(t), so du=cos(t)dt,

sqrt(1-u^2)=sqrt(cos^2(t))=cos(t)

0.5*integral(du/sqrt(1-u^2))=

=0.5*integral(dt)=0.5*t+C=

=0.5*arcsin(u)+C=

=0.5*arcsin(x^2)+C.