## bettyboop8904 can someone help me with integrals? equation below one year ago one year ago

1. bettyboop8904

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

2. khoala4pham

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. bettyboop8904

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

4. bettyboop8904

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

5. khoala4pham

one half that...chain rule!

6. bettyboop8904

so it would be 2tan(2t)?

7. bettyboop8904

= (

8. khoala4pham

no no. 1/2 * tan2t

9. bettyboop8904

is that a law? = (

10. khoala4pham

? 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. bettyboop8904

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. bettyboop8904

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

13. khoala4pham

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. bettyboop8904

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

15. khoala4pham

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

16. bettyboop8904

mine seems to be slower than a snails' = (