antiderivative of square root of x(1-x)

[gluck]

the problem is square root of x(1-x)

I don't know if I am on the right track or not.

square root of x(1-x)
=
square root of (x-x^2)
=
(x-x^2)^1/2
=
1/2(x-x^2)^-1/2
=
(1-2x)/2 square root of x-x^2

 

[skeeter]

the title of your post says "antiderivative"

you took the derivative. if that was your intent, the derivative is correct.

 

[gluck]

no it is suppossed to be the anti. is it
sqrt x * sqrt (1-x)
=
x^1/2 * (1-x)^1/2
=
2/3x^3/2*(1-x)^1/2

I really get lost after that.

 

[pka]

Go to this site: http://integrals.wolfram.com/index.jsp
Enter Sqrt[x]Sqrt[1-x] and click compute.
You will see an answer. It seems that we need to use parts to get that answer.

 

[galactus]

Here's a thougth about another approach.

\(\displaystyle \L\\\int\sqrt{x-x^{2}}dx\)

Let \(\displaystyle \L\\u=\frac{2x-1}{2}, \;\ x=\frac{2u+1}{2}, \;\ dx=du\)

Making the subs gives you:

\(\displaystyle \L\\\frac{1}{2}\int\sqrt{1-4u^{2}}du\)

Now, if you let \(\displaystyle \L\\u=\frac{1}{2}sin{\theta}, \;\ du=\frac{1}{2}cos{\theta}d{\theta}\)

and make the subs you get:

\(\displaystyle \L\\\frac{1}{4}\int{cos^{2}{\theta}}d{\theta}\)

Now continue and don't forget to make the resubs.

Just another way of looking at it if you wish.

 

[Deleted member 4993]

Similar to Galactus's way:

x = [sin(u)]^2

dx = 2* sin(u)*cos(u)

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

then

sqrt[x(1-x)]dx

= 1/2 * [sin(2u)]^2 du

= 1/4 * [1 - cos(4u)] du

Now you can continue....