# Integration by substitution help

• Jun 16th 2010, 01:26 PM
Tweety
Integration by substitution help
Use the substitution x = sin\theta to find the exact value of;

$\displaystyle \int^{\frac{1}{2}}_0 \frac{1}{( {1-x^{2}})^\frac{3}{2}}$

$\displaystyle \int \frac{1}{(1-sin^{2}\theta)^{\frac{3}{2}}}$

$\displaystyle \frac{dx}{d\theta} = cos\theta$

so when I put this all together I get ;

$\displaystyle \int \frac{1}{(cos^{2}\theta)}^{\frac{3}{2}} cos\theta d\theta$

and there does not see to be much I can do from here, cant seem to cancel the cosines,

and the 3/2 power is suppose to be at the bottom, of the fraction. Cant seem to do it on latex
• Jun 17th 2010, 12:42 PM
Ackbeet
Try re-posting this one for now without LaTeX.
• Jun 17th 2010, 01:33 PM
Ackbeet
Or with LaTeX, now. You have to manually put in the
Code:

$$ tags around the LaTeX source code, I think. • Jun 17th 2010, 06:15 PM TheCoffeeMachine Here... unfortunately, though, you may have to wait for the latex to work. Did you forget that (x^m)^(n/p) = x^{(mn)/p}, i.e. {(cosx)^2}^(3/2) = (cosx)^{(3*2)/2} = cos^3{x}? (Giggle) Full solution here, but you may have to wait for the latex to work. (Thinking) We have \displaystyle \int_{0}^{\frac{1}{2}}\dfrac{1}{\left(1-x^2\right)^{\frac{3}{2}}}\;{dx} If we let (as suggested) \displaystyle x = \sin{\theta}, then \displaystyle \dfrac{dx}{d\theta} = \cos{\theta} \Rightarrow dx = d\theta\cos{\theta}. When \displaystyle x = \dfrac{1}{2}, \displaystyle \sin\theta = \dfrac{1}{2} \Rightarrow \theta = \dfrac{\pi}{6}; similarly, when \displaystyle x = 0, \displaystyle \sin{\theta} = 0 \Rightarrow \theta = 0. Then we have \displaystyle \int_{0}^{\frac{1}{2}}\dfrac{1}{\left(1-x^2\right)^{\frac{3}{2}}}\;{dx} = \int_{0}^{\frac{\pi}{6}}\dfrac{\cos{\theta}}{\left (1-\sin^2{x}\right)^{\frac{3}{2}}}\;{d\theta} \displaystyle = \int_{0}^{\frac{\pi}{6}}\dfrac{\cos{\theta}}{\left (\cos^2{\theta}\right)^{\frac{3}{2}}}\;{d\theta} = \int_{0}^{\frac{\pi}{6}}\dfrac{\cos{\theta}}{\left (\cos^3{\theta}\right)}\;{d\theta}  \displaystyle = \int_{0}^{\frac{\pi}{6}}\dfrac{1}{\left(\cos^2{\th eta}\right)}\;{d\theta} = \int_{0}^{\frac{\pi}{6}}\sec^2{x}\;{d\theta} = \bigg[\tan{x}\bigg]_{0}^{\frac{\pi}{6}} = \tan\dfrac{\pi}{6} = \dfrac{1}{\sqrt{3}} • Jun 17th 2010, 06:54 PM undefined Quote: Originally Posted by TheCoffeeMachine Full solution here, but you may have to wait for the latex to work. (Thinking) Used this page to make an image, useful for now, click on image for full size: Attachment 17894 For fast results, used search-and-replace [tex] -> }$$ -> \text{

and then check the beginning and end to make sure no missing or extra \text{ or }
• Jun 17th 2010, 07:04 PM
simplependulum
Here is a trick , substitute x = 1/t we have dx = -dt/t^2 and it becomes

int_{oo}^{2} (-dt/t^2)/( 1 - 1/t^2 )^(3/2)

= int_{2}^{oo} [dt/t^2(t^3)] /( t^2 -1 )^(3/2)

= int_{2}^{oo} (t dt)(t^2 - 1)^(-3/2)

= [ (t^2-1)^(-1/2) (-2) / 2 ]_{2}^{oo}

= - [ 1/sqrt(t^2-1) ]_{2}^{oo}

= - 0 + 1/sqrt(2^2-1) = 1/sqrt(3)
• Jun 17th 2010, 07:39 PM
TheCoffeeMachine
Attachment 17895
Quote:

Originally Posted by undefined
Used this page to make an image, useful for now, click on image for full size:

Attachment 17894

For fast results, used search-and-replace

$$-> }$$ -> \text{

and then check the beginning and end to make sure no missing or extra \text{ or }

Cool, thanks! Recreated the picture, with corrections of typos that I've (no wonder) made earlier:

http://www.mathhelpforum.com/math-he...6&d=1276832980
• Jun 17th 2010, 07:51 PM
undefined
Quote:

Originally Posted by TheCoffeeMachine
Cool, thanks! Recreated the picture, with corrections of typos that I've (no wonder) made earlier:

You're welcome!

How did you get the inline image to appear full size? I see that the code is

[IMG]http://www.mathhelpforum.com/math-help/attachment.php?attachmentid=17895&d=1276832168[/IMG]

but what steps did you use to produce it?
• Jun 17th 2010, 08:06 PM
TheCoffeeMachine
Quote:

Originally Posted by undefined
You're welcome!

How did you get the inline image to appear full size? I see that the code is

[IMG]http://www.mathhelpforum.com/math-help/attachment.php?attachmentid=17895&d=1276832168[/IMG]

but what steps did you use to produce it?

After uploading, click on the attachment (where it says Attached Thumbnails), and then it will take you to the image url. (When you click to view the image you posted earlier in full size, clicking again will give you the URL).

EDIT: Just realised that might not have been clear.

After uploading, I clicked 'preview post' and clicked on the attachment, then it took me to here. I copied http://www.mathhelpforum.com/math-he...5&d=1276832168 from the address bar and went back to replying, clicked on the image icon http://www.mathhelpforum.com/math-he...nsertimage.png in the tool bar, and pasted it on the little box that appears, and the picture appeared full.