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bettyboop8904Best ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}tsec ^{2}2t\]
 one year ago

khoala4phamBest ResponseYou've already chosen the best response.0
like your last problem, this is another integration by parts. How do we know? because sec^2 is ordinarily integrable but it's multiplied by t. Here, we let u = t, and dv = sec^2 (t). du = dt and v = tan(t) with this information, you should be able to solve.
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
lol i literally just figured it out right when you posted that lol ty though = )
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
one question though dv=sec^2 (2t) would be v=tan(2t) right?
 one year ago

khoala4phamBest ResponseYou've already chosen the best response.0
one half that...chain rule!
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
so it would be 2tan(2t)?
 one year ago

khoala4phamBest ResponseYou've already chosen the best response.0
no no. 1/2 * tan2t
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
is that a law? = (
 one year ago

khoala4phamBest ResponseYou've already chosen the best response.0
? well, the chain rule is not a "law" either you believe in it or you don't. If you don't then most mathematicians just won't work with you. :P
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
so wait everytime you have a trig sign with f(x) and you take the integral its that trig sign with f(x) multiplied by 1/f(x) ?
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
i'm sorry maybe i only remember the chain rule for derivatives... not integrals...
 one year ago

khoala4phamBest ResponseYou've already chosen the best response.0
Well I mean it's a simple consequence of u substitution is what I mean. Suppose you wanted to integrate cos2t. you technically only know the formula for the integral of cost. But by being clever, you let u = 2t. du = 2dt. thus dt = (1/2)du. AGAIN I MUST REMIND YOU THESE ARE NOT FRACTIONS but they work out that way for these general cases. So (integral)cos2tdt = (integral) (1/2) cosu du. Why? because originally, we integrated with respect to dt. So we must change everything that is in terms of to to everything in terms of u. That is why this integral evaluates to 1/2 tan(2t)
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
ok thank you i'm sorry i suck at this = (
 one year ago

khoala4phamBest ResponseYou've already chosen the best response.0
Hey no problem. It's always good to learn. We each have our own pace.
 one year ago

bettyboop8904Best ResponseYou've already chosen the best response.0
mine seems to be slower than a snails' = (
 one year ago
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