# Rationalizing Substitutions

• Jul 14th 2010, 04:21 PM
katchat64
Rationalizing Substitutions
[IMG]file:///C:/Users/Karina/AppData/Local/Temp/msohtmlclip1/01/clip_image002.gif[/IMG]Attachment 18198

This is the equation I am having trouble with. I think that 2sin(u) is what I need to use for x. Once I substitute everything in and begin to do all the steps, I get stuck. I'll show what I have so far. If I done something wrong please feel free to correct me. I also know what my answer should be, but I don't want just the answer I need help understanding how to get the answer.

Attachment 18200

[IMG]file:///C:/Users/Karina/AppData/Local/Temp/msohtmlclip1/01/clip_image006.gif[/IMG]
• Jul 14th 2010, 04:51 PM
apcalculus
Is WolframAlpha acceptable? If it is, be sure to click on 'Show Steps'.

http://www.wolframalpha.com/input/?i=integrate+(sqrt(4-x^2)/x)

Good luck!
• Jul 14th 2010, 05:34 PM
Prove It
You got to

$\displaystyle 2\int{\frac{\cos^2{u}}{\sin{u}}\,du}$ alright.

$\displaystyle = 2\int{\frac{1 - \sin^2{u}}{\sin{u}}\,du}$

$\displaystyle = 2\int{\frac{1}{\sin{u}} - \frac{\sin^2{u}}{\sin{u}}\,du}$

$\displaystyle = 2\int{\csc{u} - \sin{u}\,du}$

$\displaystyle = 2\left[-\ln{(\cot{u} + \csc{u})} + \cos{u} + C\right]$.

Now try to convert it back to a function of $\displaystyle x$.
• Jul 14th 2010, 05:42 PM
katchat64
Can you use the half angle formula for cos^2x? Or would that make the problem harder to evaluate? cos^2x = (1+cos^2x)/2
• Jul 14th 2010, 05:47 PM
Prove It
First of all

$\displaystyle \cos^2{x}$ does NOT equal $\displaystyle \frac{1 + \cos^2{x}}{2}$, it's actually $\displaystyle \frac{1 + \cos{2x}}{2}$.

Second, no you can't. Since you're trying to simplify the integrand, you would want the numerator to be something with the sine function, so that you can cancel.