# Thread: evaluating integrals and using substitution

1. ## evaluating integrals and using substitution

Hi I was wondering if someone would be able to help me evaluate \int \sqrt{x/1-x} with limits of 0 and 0.5, by using the substitution of x=sin2t - (thats sin squared t)

2. Originally Posted by ella745
Hi I was wondering if someone would be able to help me evaluate \int \sqrt{x/1-x} with limits of 0 and 0.5, by using the substitution of x=sin2t - (thats sin squared t)

$\displaystyle \displaystyle\int_0^{1/2}\sqrt{\dfrac{x}{1-x}}\;dx=\displaystyle\int_0^{\pi/4}\sqrt{\dfrac{\sin^2 t}{\cos ^2 t}}\sin 2t\;dt=\ldots=2\displaystyle\int_0^{\pi/4}\sin^2 t\;dt=\ldots$

3. Originally Posted by ella745
Hi I was wondering if someone would be able to help me evaluate \int \sqrt{x/1-x} with limits of 0 and 0.5, by using the substitution of x=sin2t - (thats sin squared t)
If x = sin^2(t), then dx/dt = 2sin(t)cos(t), or dx = 2sin(t)cos(t)dt.

So int{sqrt[x/(1 - x)]dx} = int{sqrt[sin^2(t)/(1 - sin^2(t))]2sin(t)cos(t)dt}

= int{2sqrt[sin^2(t)/cos^2(t)]sin(t)cos(t)dt}

= int{2[sin(t)/cos(t)]sin(t)cos(t)dt}

= int{2sin^2(t)dt}.

Can you go from here?

4. thank you very much. are you able to fill in the middle steps? if not that is ok!