## anonymous 4 years ago Need help with another intergral. Never really done one like this before. tan^6(3y)sec^2(3y)dy

1. anonymous

fo u know c++

2. anonymous

$\int\limits_{?}^{?}\tan ^6\left( 3y \right)\sec ^2(3y)dy$

3. anonymous

I think I need to do u=tan(3y)? Maybe not even close? ,

4. anonymous

or u=tanx? So du=sec^2x?

5. anonymous

Anyone have any clue?

6. anonymous

turing test ..do u knw c++?

7. TuringTest

yes, that will work, but it's$u=\tan(3y)$$du=3\sec^2(3y)$remember to use the chain rule all the way through @shiv no, sorry

8. TuringTest

sorry$u=\tan(3y)$$du=3\sec^2(3y)dy$can't forget the dy! very important

9. anonymous

so would it be $1/3\int\limits_{?}^{?}\tan(u)^6dy?$

10. TuringTest

no,$u=\tan(3y)$so after the sub there should be no tangent left

11. anonymous

oh... its $1/3\int\limits_{?}^{?}u^6du$

12. TuringTest

yep :D

13. anonymous

you are a life saver. Can I just ask you one more quick here

14. TuringTest

sure but I'm headed for dinner so gotta be quick

15. anonymous

ill try. $\int\limits_{-1}^{-1/2}t^-2 \sin^2(1+1/2)dt sorry its t \to the -2 wont show up though$

16. anonymous

Any direction helps. Do i have to set something u ,, then eventually change the limits of integration? I never know when to do that and feel this could be one..

17. TuringTest

$\int t^{-2}\sin^2?dt$sorry, what is the integrand exactly?

18. anonymous

the end is sin^2(1+1/t)

19. anonymous

that 2 was wrong. you got the rest right. evaluated from -1 to -.5

20. TuringTest

$\int t^{-2}\sin^2(1+\frac1t)dt$yes?

21. anonymous

yes

22. anonymous

evaluated from -1 to -1/2 if that changes anything

23. TuringTest

$u=1+\frac1t$you don't have to change the bounds if you turn it back to t at the end, but if you want to evaluate it in terms of u you need to do this:$u_1=1+\frac1{-1}=0$$u_2=1+\frac1{-1/2}=-1$so those are your new limits (if you want to evaluate in terms of u the differential is$du=-t^{-2}dt$so the integral is$-\int_{0}^{-1}\sin udu=\int_{-1}^{0}\sin u du$Ok gotta eat, good luck!

24. anonymous

Thanks a lot! I appreicate it