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OpenStudy (christos):
Trigonometric substitutions,
can you please help me with these?
https://www.dropbox.com/s/agmk6x5vvm0wne0/Screenshot%202013-10-25%2022.53.06.jpg
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OpenStudy (anonymous):
have a look...
OpenStudy (christos):
why do I feel that this is totally irrelevant :o
OpenStudy (christos):
I know there is a method to solve these, but unfortunately I don't remember it, there is a specific method.
OpenStudy (anonymous):
top of the third page...
OpenStudy (christos):
Ok I found it, wait , could you tell me the first three steps of lets say, 1 (a) ? I have like, completely forgotten it, all I need is to recall it.
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OpenStudy (christos):
like in the top of the third page, could we separate it in smaller steps ?
OpenStudy (anonymous):
1a. Let x = a sin theta
OpenStudy (christos):
ok
OpenStudy (anonymous):
\[\sqrt{a^2-x^2}\Rightarrow \sqrt{a^2-a^2\sin^2\theta}= acos\theta\]
since x = a sin theta, dx = a cos theta
OpenStudy (christos):
so I am in the step of int[sqrt(2^2 - sin^2(theta)aces(theta)
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OpenStudy (christos):
cos*
OpenStudy (christos):
Now what ? @pgpilot326
OpenStudy (anonymous):
why do I feel that this is totally irrelevant :o
OpenStudy (christos):
I am a beginner.
OpenStudy (christos):
didn't mean to hurt you ;p
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