## anonymous 5 years ago Someone please help me real quick. Im going to post the question now

1. anonymous

given $\cos(\sqrt{t})/\sqrt{t}$ If l let u = sqt(t) then does that mean it applies to both cos funtion and root funtion, or just the inner most sqrt inside cos????????

2. anonymous

both..........

3. anonymous

that doesnt help me for taking the indefinite integral though

4. anonymous

because u have given new value to sqrt (t) no matter where it is

5. anonymous

u might need to post your question to have opinions on that problem

6. anonymous

$\int\limits_{}^{}\cos(\sqrt{t})/\sqrt{t}$

7. anonymous

You need to consider $\frac{\mathbb{d}u}{\mathbb{d}x}$

8. anonymous

/dt*

9. anonymous

try sqrt(t)= z

10. anonymous

I was trying to get ride of the root t in the botom with $dt = 2\sqrt{t} du$

11. anonymous

You need to substitute in u for root t there aswell..

12. anonymous

after substituting z=t^1/2 u give get as below integration sign ( cosz * 2z^2 dt )

13. anonymous

then u can use integration by parts keeping z^2 as first term and cosz as second

14. anonymous

let me show you what im trying

15. anonymous

let u = $\sqrt{t}$ then $du = 1/(2\sqrt{t})dt$ $dt = 2\sqrt{t}du$ then i get $\int\limits_{}^{} \cos(u)/u 2\sqrt{t}du$

16. anonymous

is this wrong?

17. anonymous

change t^1/2 to u because now we cant have t in new eqation, now we are dealing with u only

18. anonymous

oooooooooooo

19. anonymous

got it lol

20. anonymous

u r on right track ..well done

21. anonymous

bang !!! cheers!!! good luck with other problems too !!

22. anonymous

thank you so much!

23. anonymous

keep up the gud work !!!