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

1. anonymous

$\int\limits_{}^{}tsec ^{2}2t$

2. anonymous

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.

3. anonymous

lol i literally just figured it out right when you posted that lol ty though = )

4. anonymous

one question though dv=sec^2 (2t) would be v=tan(2t) right?

5. anonymous

one half that...chain rule!

6. anonymous

so it would be 2tan(2t)?

7. anonymous

= (

8. anonymous

no no. 1/2 * tan2t

9. anonymous

is that a law? = (

10. anonymous

? 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

11. anonymous

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) ?

12. anonymous

i'm sorry maybe i only remember the chain rule for derivatives... not integrals...

13. anonymous

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)

14. anonymous

ok thank you i'm sorry i suck at this = (

15. anonymous

Hey no problem. It's always good to learn. We each have our own pace.

16. anonymous

mine seems to be slower than a snails' = (