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bettyboop8904
 3 years ago
can someone help me with integrals? equation below
bettyboop8904
 3 years ago
can someone help me with integrals? equation below

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bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}tsec ^{2}2t\]

khoala4pham
 3 years ago
Best ResponseYou've already chosen the best response.0like 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.

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0lol i literally just figured it out right when you posted that lol ty though = )

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0one question though dv=sec^2 (2t) would be v=tan(2t) right?

khoala4pham
 3 years ago
Best ResponseYou've already chosen the best response.0one half that...chain rule!

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0so it would be 2tan(2t)?

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0is that a law? = (

khoala4pham
 3 years ago
Best 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

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0so 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) ?

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0i'm sorry maybe i only remember the chain rule for derivatives... not integrals...

khoala4pham
 3 years ago
Best ResponseYou've already chosen the best response.0Well 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)

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0ok thank you i'm sorry i suck at this = (

khoala4pham
 3 years ago
Best ResponseYou've already chosen the best response.0Hey no problem. It's always good to learn. We each have our own pace.

bettyboop8904
 3 years ago
Best ResponseYou've already chosen the best response.0mine seems to be slower than a snails' = (
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