# hi i need to integrate this.

• May 19th 2009, 04:40 PM
mbbx5va2
hi i need to integrate this.
Hi i'm in a bit of a hurry, any help would be great please see the attachment.
• May 19th 2009, 05:18 PM
e^(i*pi)
Quote:

Originally Posted by mbbx5va2
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
• May 19th 2009, 05:24 PM
Nacho
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?
• May 19th 2009, 07:28 PM
Krizalid
put $\displaystyle x=\frac{1}{1+t^{2}}$ and your integral becomes $\displaystyle \int_{0}^{\infty }{\frac{2}{1+t^{2}}\,dt}=\pi .$
• May 19th 2009, 10:28 PM
matheagle
Quote:

Originally Posted by e^(i*pi)
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

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.
• May 20th 2009, 10:32 AM
e^(i*pi)
Quote:

Originally Posted by matheagle
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.

-facepalm-

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