Definite Integral of $\int_0^1\frac{dx}{\sqrt {x(1-x)}}$ We have to calculate value of the following integral :
$$\int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \qquad \qquad (2)$$
What i've done for (2) :
\begin{align}
 & = \int_0^1\cfrac{dx}{\sqrt {x(1-x)}} \\
 & = \int_0^1\cfrac{dx}{\sqrt {x-x^2}} \\ 
 & = \int_0^1\cfrac{dx}{\sqrt {(x^2-x+\frac 14)-\frac 14 }} \\
 & = \int_0^1\cfrac{dx}{\sqrt {(x-\frac 12)^2-(\frac 12)^2 }} \\ 
 & = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 12\sec \theta)^2-(\frac 12)^2 }}   I\ used\ trigonometric\ substitution \ u=a\sec \theta, by \ it's \ form \ u^2-a^2    \\
 & = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {(\frac 14\sec^2 \theta)-\frac 14 }}  \\
 & = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 14(\sec ^2\theta-1)}} \ using \\tan^2\theta=\sec^2\theta-1 \\
 & = \cfrac {1}{2}\int_0^1\cfrac{\sec \theta \tan \theta \ d\theta}{\sqrt {\frac 12(\sqrt{\tan^2\theta) }}} \\
 & = \int_0^1\sec\theta d\theta = \sec\theta \tan \theta |_0^1                      \\
\end{align}
But i got problems calculating $\theta$ value, using trigonometric substitution, any help?
 A: Are you sure that you got the correct antiderivative for the integral of the secant function? The correct indefinite integral of secant is $\int\sec{x}=\ln{|\tan{x}+\sec{x}|}+C$. Thus:
$$
\int_{0}^{1}\sec{\theta}\,d\theta=\ln{|\tan{\theta}+\sec{\theta}|}\bigg|_{0}^{1}.
$$
EDIT:
Also note that:
$$
x-x^2=-(x^2-x)=-\left(x^2-x+\frac{1}{4}-\frac{1}{4}\right)=\\
-\left(\left[x-\frac{1}{2}\right]^2-\frac{1}{4}\right)=
\frac{1}{4}-\left(x-\frac{1}{2}\right)^2.
$$
So, I think the substitution that you should be using would be:
$$
x=\frac{1}{2}\sin{\theta}+\frac{1}{2},\\
dx=\frac{1}{2}\cos{\theta}\,d\theta,\\
x=\frac{1}{2}\sin{\theta}+\frac{1}{2}\implies\theta=\arcsin{(2x-1)},\ -\frac{\pi}{2}\le\theta\le\frac{\pi}{2}.
$$
Putting it all together, you get the following:
$$
\int\frac{1}{\sqrt{x-x^2}}\,dx=\int\frac{1}{\sqrt{\frac{1}{4}-\left(\frac{1}{2}\sin{\theta}+\frac{1}{2}-\frac{1}{2}\right)^2}}\frac{1}{2}\cos{\theta}\,d\theta=\\
\int\frac{1}{\frac{1}{2}\sqrt{1-\sin^2{\theta}}}\frac{1}{2}\cos{\theta}\,d\theta=
\int\frac{\cos{\theta}}{|\cos{\theta}|}\,d\theta=
\int\frac{\cos{\theta}}{\cos{\theta}}\,d\theta=\\
\int\,d\theta=\theta+C=
\arcsin{(2x-1)}+C.\\
\int_0^1\frac{1}{\sqrt {x(1-x)}}\,dx=\arcsin{(2x-1)}\bigg|_0^1=\\
\arcsin{(1)}-\arcsin{(-1)}=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi.
$$
Wolfram Alpha check.
A: Letting $x=\sin^2 \theta$ yields
$$
\begin{aligned}
\int_0^1 \frac{d x}{\sqrt{x(1-x)}} &=\int_0^{\frac{\pi}{2}} \frac{2 \sin \theta \cos \theta d \theta}{\sin \theta \cos \theta} =[2 \theta]_0^{\frac{\pi}{2}} =\pi
\end{aligned}
$$
A: You must convert the limits to be in terms of $\theta$. When $x=0,1$, what is $\theta$?
Also if you want another method consider partial fractions
A: As $0\le x\le1$ 
WLOG $x=\sin^2t;0\le t\le\dfrac\pi2,dx=?$
So $\sqrt x=+\sin t,\sqrt{1-x}=?$
Alternatively $4x(1-x)=1-(2x-1)^2$
Set $2x-1=\sin y$ or $\cos y$
Observe that $2x-1=-\cos2t$ in the first method
