what is the integral of x / sqrt 1 - x^4 ?

Answer 1

Q. what is the integral of x / sq rt 1 - x^4
Ans:
put x^2 = sin A
Differentiate
2x dx = cos A dA 0r x dx = 1/2 cos A dA
1/2 integral( cos A/ cos A )dA because sqrt (1 - Sin^2A)
!/2 integral dA= 1/2 A+C=1/2 sin^-1x^2 +C

Answer 2

Write your integral as 1/2 int[ 2x dx/ sqrt(1-x^4)].
Now let u = x^2, du = 2x dx
The rest is simple!

Answer 3

Let u =x^2 so 1/2 du=x dx.

Then,

\int x/sqrt(1-x^4) dx=/2 \int 1/sqrt(1-u^2) du=1/2 arcsin(u)+C
=1/2 arcsin(x^2)+C

Note 1: If you forget that \int 1/sqrt(1-u^2) du= arcsin(u)+C you can either do a trig subsitution or derive it using inverse functions and the chain rule:

Using inverse functions and the chain rule,

sin( arcsin(u))=u

so

d/du (sin( arcsin(u))) d/du (u)

Now apply the chain rule:

cos(arcsin(u)) * d/du(arcsin(u))=1

Dividie

d/du(arcsin(u))=
1/cos(arcsin(u))

d/du(arcsin(u))=
1/sqrt(1-u^2)

Note 2:

[cos(arcsin(u))]^2+
[sin(arcsin(u))]^2=1

[cos(arcsin(u))]^2+u^2=1

[cos(arcsin(u))]^2=1-u^2

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

Answer 4

use trig sub. Its tough to demostrate in text.

think of it a a trianlge: http://img.photobucket.com/albums/v613/HL2photo/trigsub.jpg
the x on the right leg should be x^2
the other leg shoudl be sqrt(1-x^4)

Then let sinO=x (pretend O is theta)
So -cosO dO = dx

then plug in

int x dx/sqrt(1-x^4)
int -cos(O)sin(O) dO/sqrt(1-sin^4(O))
int -cos(O)sin(O) dO/sqrt(cos^4(O))
int -cos(O)sin(O) dO/cos^2(O)
int -sin(O) dO/cos(O)
then u-sub it.
u=cos(O)
du=-sin(O) dO
int du/u
ln|u|+C
ln|cos(O)|+C

now look back at the triangle to solve for cos(O)
so the final answer is
ln(sqrt(1-x^4))+C