Trigonometric Integrals - Part 6 of 6
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Uploader Comments (patrickJMT)
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All Comments (107)
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Hey Patrick! I believe you made a mistake in the first problem.. you have two minuses in your answer so you should get a positive 1/2cos2x! Anway thanks for posting these videos up they were extremely helpful! :)
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@lintingche2012 if you let sinx = u, then it follows cosxdx = du, but u dnt have cosx in the numerator.. if the given is cotxdx the answer will be ln(sinx)
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am lost in this part 6 patrick
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Watch Patrick videos
Drink lots of coffee
Study study study
Profit!
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So this last part of trig integrals is basically for any problems that you wouldn't be able to approach using the other five steps without doing some trig identity or algebra first?
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thanks patrick!!!
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I don't go to classes, watch your videos and get good grades on my tests :) thank u!!
VKM7md 1 week ago
@VKM7md :) go to class.
patrickJMT 1 week ago
Patrick
you make calculus 2 look like 1+1=2!
thanks
ricky23i 1 month ago 5
@ricky23i you are very welcome :)
patrickJMT 4 weeks ago
For the integration of Csc(x)
why can't I just differentiate it as 1/sinx
than let u=sinx
and integral (1/u) = ln(u)
ln(sinx)?
lintingche2012 3 months ago
@lintingche2012 what about the du? you should rewatch u-substitutions. according to what you are saying, integral of 1/(anything) = ln(anything). take the derivative of ln(sinx) and see if you get csc(x); if not, something is wrong (and it is!)
patrickJMT 2 months ago 3