# Thread: hi i need to integrate this.

1. ## hi i need to integrate this.

Hi i'm in a bit of a hurry, any help would be great please see the attachment.

2. Originally Posted by mbbx5va2
Hi i'm in a bit of a hurry, any help would be great please see the attachment.
Question is: $\int^1_0 \frac{1}{\sqrt{x(1-x)}} dx$

Use $x = sin^2(\theta)$ to get $\int^{\frac{\pi}{2}}_0 1 d\theta$

$x = sin^2(\theta)$ then our denominator becomes: $\sqrt{sin^2(\theta)(1-sin^2(\theta))} = \sqrt{sin^2(\theta)cos^2(\theta)}$ $= sin(\theta)cos(\theta)$

We can also use our substitution to change the limits:
$1 = sin^2(\theta) \: , \: \theta= \frac{pi}{2}$

$0 = sin^2(\theta) \: , \: \theta = 0$

$\int^{\frac{\pi}{2}}_0 \frac{1}{sin(\theta)cos(\theta)}d\theta$

Not sure where the 1 came from but that integral can be solved using integration by parts

3. Maybe you should work a bit: $x-x^2$

That seem an arcsin, then:

$\dfrac{1}{4} + \left( -\dfrac{1}{4}+x-x^2 \right)=\dfrac{1}{4} - \left( x^2-x+\dfrac{1}{4}\right)$ $=\left(\dfrac{1}{2} \right)^2-\left(x-\dfrac{1}{2} \right)^2$ $=
\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?

4. put $x=\frac{1}{1+t^{2}}$ and your integral becomes $\int_{0}^{\infty }{\frac{2}{1+t^{2}}\,dt}=\pi .$

5. Originally Posted by e^(i*pi)
Question is: $\int^1_0 \frac{1}{\sqrt{x(1-x)}} dx$

Use $x = sin^2(\theta)$ to get $\int^{\frac{\pi}{2}}_0 1 d\theta$

$x = sin^2(\theta)$ then our denominator becomes: $\sqrt{sin^2(\theta)(1-sin^2(\theta))} = \sqrt{sin^2(\theta)cos^2(\theta)}$ $= sin(\theta)cos(\theta)$

We can also use our substitution to change the limits:
$1 = sin^2(\theta) \: , \: \theta= \frac{pi}{2}$

$0 = sin^2(\theta) \: , \: \theta = 0$

$\int^{\frac{\pi}{2}}_0 \frac{1}{sin(\theta)cos(\theta)}d\theta$

Not sure where the 1 came from but that integral can be solved using integration by parts
This looks fishy... $dx\ne d\theta$.

If you use $x = \sin^2\theta$, then $dx = 2\sin\theta\cos\theta d\theta$ which will give you

$\int^{\frac{\pi}{2}}_0 2 d\theta=\pi$ and that's agrees with another imposter's post.

6. Originally Posted by matheagle
This looks fishy... $dx\ne d\theta$.

If you use $x = \sin^2\theta$, then $dx = 2\sin\theta\cos\theta d\theta$ which will give you

$\int^{\frac{\pi}{2}}_0 2 d\theta=\pi$ and that's agrees with another imposter's post.
-facepalm-

how did I not see that? Thanks for pointing that out