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Mathematics 35 Online
OpenStudy (anonymous):

Trig substitution I have the indefinite integral int 1/(4-z^2)^3/2 dz but I'm confused how to draw a triangle to use trig substitution (because I'm pretty sure this is a trig substitution problem) Thanks

OpenStudy (anonymous):

I'm honestly not sure what the integral is. Is there a fraction inside a fraction? Try using the "Equation" toolbox to make it more clear

OpenStudy (anonymous):

You can't use it when typing a question and if I would have typed in in the comments then it shows up as "answered" so then people don't answer at all. The integral is \[\int\limits_{}^{}\frac{ 1 }{ (4-z^2)^\frac{ 3 }{ 2 } }dz\]

OpenStudy (anonymous):

Ohh I see. Next time put brackets around the (3/2), I would have gotten it then. I'm not sure how to do it with drawing triangles, but I would substitute z=2cos^2u

OpenStudy (anonymous):

It's either that or z=2sin^2u, I haven't done cal 2 in a while :P

zepdrix (zepdrix):

\[z=2\sin \theta\]\[dz=2\cos \theta \; d \theta\]Changing the integral to,\[\int\limits \frac{dz}{(4-z^2)^{3/2}} \qquad \rightarrow \qquad \int\limits \frac{2\cos \; d \theta}{(4-4\sin^2 \theta)^{3/2}}\]

zepdrix (zepdrix):

Sine or cosine will work just fine :) Lemme see if i can draw out the triangle for you :O it's a little tricky since im on a laptop :c but ill try.

zepdrix (zepdrix):

|dw:1355006887680:dw|Understand how I set that up? :O

zepdrix (zepdrix):

From there you can easily solve for the missing side using the Pythagorean Theorem.

OpenStudy (anonymous):

so basically all I needed to do was find something that fit \[4\cos^2\theta+4\sin^2\theta=4\]? and substitute the whole \[(4-z^2)\] for \[4\cos^2\theta\]?

OpenStudy (anonymous):

my most simplified integral is \[\int\limits_{}^{}\frac{ 1 }{ 4\cos^2\theta }d \theta\]

zepdrix (zepdrix):

Hmm yah I think that looks right :D

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