## anonymous 5 years ago if 0 < k <pi, then 0 (integral) k cos(2x) dx = 1/2 when k=?

1. anonymous

use equation option

2. anonymous

$\int\limits_{0}^{k} \cos(2x) dx =1/2$ thats what i meant

3. anonymous

[sin(2k)/2]-sin(0/2)=sin(2k)/2=1/2 sin(2k)=1 2k=[\pi/2\] so k=$\pi/4$

4. anonymous

why do you divide by 2?

5. anonymous

as you have 2x

6. anonymous

in more detail put 2x=t..........(1) now differentiate both sides 2dx=dt dx=dt/2 and limits will change like this t=0 at x=0 .......from (1) t=2k

7. anonymous

now solve interms of t

8. anonymous

t is a random variable or what does it stand for?

9. anonymous

random

10. anonymous

sorry i dont get it still, i guess cause thats not how they have taught us over here...um are you showing me the antiderivative?

11. anonymous

what's ur country name?

12. anonymous

United States of America

13. anonymous

k i was showing u the antiderivative

14. anonymous

ohh ok.. i was told to add one to the power and divide by the new power, is that what you did?

15. anonymous

it in the case of polynomial i mean sometjing like....x^2 or x^100

16. anonymous

um ok?

17. anonymous

if you have sin (polynomial of degreee 1) or cos(polynomial of degree 1) then simply do integretion 4 sin(x) or cos(x) then divide it by integretion of polynomial

18. anonymous

thats why you divided by 2 since it was in the polynomial?

19. anonymous

sorry its late and im a sort of slow now

20. anonymous

21. anonymous

lol

22. anonymous

23. anonymous

a friend of mine in calculus is last named patel he is from india, in any way related? i know big world but we found a cousin of his in a math competition lol

24. anonymous

no i don't know him but i m also from INDIA

25. anonymous

i guess its a popular last name from over there

26. anonymous

yep

27. anonymous

what's ur full name

28. anonymous

mine?

29. anonymous

yeah!

30. anonymous

benjamin vasquez

31. anonymous

got it any more q benjamin

32. anonymous

nope! thank you so much