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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.
lol i literally just figured it out right when you posted that lol ty though = )
one question though dv=sec^2 (2t) would be v=tan(2t) right?
one half that...chain rule!
so it would be 2tan(2t)?
no no. 1/2 * tan2t
is that a law? = (
? 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
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) ?
i'm sorry maybe i only remember the chain rule for derivatives... not integrals...
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)
ok thank you i'm sorry i suck at this = (
Hey no problem. It's always good to learn. We each have our own pace.
mine seems to be slower than a snails' = (