Calculus II (Part 1) Integrating with Special Substitutions
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Uploader Comments (MuchoMath)
Top Comments
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This is great its so funny,how that guy always reacts.
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This is funny; but, this is a terrible way to help beginners. XD
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All Comments (20)
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Great Job Professor Pérez. Me gustó mucho su explicación. Me gustaría que fuera un poco más lento, pero como quiera es muy bueno.
Gracias,
Luis
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compare to calc 2, calc 1 is easy
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you should teach at IVC
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i dont want to take calc next semester i thought trig was hard this sucks
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could you also do denominator decomposition?
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Lol I like math enough that even though I barely understand it I liek listen to it. Make sense? Didn't think so.
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HEY!
ayershov777 1 month ago
@ayershov777 Try my faculty website for more Calculus stuff. use the URL link in my profile!
MuchoMath 1 month ago
Loved it, wish there were more :D
smokenbiskits 1 year ago
@smokenbiskits I've still a lot more work to do.
MuchoMath 1 year ago
does u= tan(x/2) for only this problem or for all problems similar to this one like 1 over some number of sinx - some number of cosx?
xxcowslayerxxx 2 years ago
This works for any rational function similar to this involving any of the six trig functions. Try integrating 1 / (sinx + tanx) dx. You should get Int(1-u^2) / 2u du = 1/2 ln|tan(x/2)| - 1/4 ln|tan^2(x/2)| +C . Next try integrating 1 / (1+sinx+cosx) dx Here you should get Int1/(u+1) du = ln|tan(x/2)| + 1| + C . Hope this helps! MuchoMath
MuchoMath 2 years ago