Hi i'm in a bit of a hurry, any help would be great please see the attachment.
Question is: $\displaystyle \int^1_0 \frac{1}{\sqrt{x(1-x)}} dx$
Use$\displaystyle x = sin^2(\theta)$ to get $\displaystyle \int^{\frac{\pi}{2}}_0 1 d\theta$
$\displaystyle x = sin^2(\theta)$ then our denominator becomes: $\displaystyle \sqrt{sin^2(\theta)(1-sin^2(\theta))} = \sqrt{sin^2(\theta)cos^2(\theta)}$ $\displaystyle = sin(\theta)cos(\theta) $
We can also use our substitution to change the limits:
$\displaystyle 1 = sin^2(\theta) \: , \: \theta= \frac{pi}{2}$
$\displaystyle 0 = sin^2(\theta) \: , \: \theta = 0$
$\displaystyle \int^{\frac{\pi}{2}}_0 \frac{1}{sin(\theta)cos(\theta)}d\theta$
Not sure where the 1 came from but that integral can be solved using integration by parts
Maybe you should work a bit: $\displaystyle x-x^2$
That seem an arcsin, then:
$\displaystyle \dfrac{1}{4} + \left( -\dfrac{1}{4}+x-x^2 \right)=\dfrac{1}{4} - \left( x^2-x+\dfrac{1}{4}\right)$ $\displaystyle =\left(\dfrac{1}{2} \right)^2-\left(x-\dfrac{1}{2} \right)^2$ $\displaystyle =
\left( {\frac{1}
{2}} \right)^2 \left\{ {1 - \left( {\frac{{x - \frac{1}
{2}}}
{{\frac{1}
{2}}}} \right)^2 } \right\}
$
can you see the substitute?
This looks fishy... $\displaystyle dx\ne d\theta$.
If you use $\displaystyle x = \sin^2\theta$, then $\displaystyle dx = 2\sin\theta\cos\theta d\theta$ which will give you
$\displaystyle \int^{\frac{\pi}{2}}_0 2 d\theta=\pi$ and that's agrees with another imposter's post.