# Thread: Rationalizing Substitutions

1. ## Rationalizing Substitutions

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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.

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2. 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!

3. You got to

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

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

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

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

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

Now try to convert it back to a function of $x$.

4. 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

5. First of all

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

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.