anonymous 5 years ago integrate sin^43x dx

1. anonymous

$\int\limits \sin^{43}(x) dx= - \cos^{43}(x)$ honestly, this is a weird question lol, so I'm not sure? ._.

2. anonymous

+ C ofcourse

3. anonymous

sstar, i don't know the answer but i'm almost sure that's not correct ><

4. anonymous

LOL >_< I know, it looks weird

5. anonymous

hmm, is it possible to let u = 43? I've never solved such thing lol

6. anonymous

meh... no...this method is funny lol

7. anonymous

sstarica the question is$\int\limits \sin ^{4}3x dx$

8. anonymous

no, you have to substitute (cos x)^2 = 1 - (sin x)^2

9. anonymous

oh that makes everything a lot better.... oneprince USE PARENTHESIS!

10. anonymous

ok chia

11. anonymous

no wait a min, isn't the sin to the power of 43?

12. anonymous

chia.. solve it

13. anonymous

was it to the power of 43? or to the power of 2? lol

14. anonymous

give me a sec

15. anonymous

it's sin(3x)^4

16. anonymous

that's the question?

17. anonymous

oh lol

18. anonymous

ugh...sub 3x for u and then use the half angle formula to get cosine and sine together, integrate

19. anonymous

sstar are you writing it out? caz i don't really want to...

20. anonymous

>_< nvm, proceed chia, I misunderstood the question lol

21. anonymous

oneprince, did you read what i wrote? is that enough?

22. anonymous

AMISTRE! you can do this! ^_^

23. amistre64

Howdy :)

24. anonymous

lol

25. anonymous

my brain is defunctioning, so I don't think I can write it down even though I took it this semester and it looks SIMPLE! would you amistre? ^_^

26. amistre64

[S] sin^4(3x) dx this is missing something lol this came from something similar to -cos^5(3x) so what do we get when we derive that? and itll tell us what to add...

27. anonymous

it's not really missing something, you can let u = 3x and solve normally like chia said

28. anonymous

if only I remember lol >_<

29. amistre64

Dx(-cos^5(3x)) = 5(3) sin^4(3x) which would be easy to solve right? So we need a "15" in front of your integral to convert it

30. amistre64

lets multiply your original problem by "1"; or rather a convenient for of "1" such as 15/15..

31. anonymous

where did you get cos^5?

32. amistre64

then we can pull out the bottom "15" and integrate the rest up like normal

33. amistre64

sin^4 comes from cos^5; or at least a version of it right?

34. anonymous

you can do this : $\int\limits (1-\cos^2(3x))^2 dx = \int\limits(1-2\cos^2u+\cos^2u)du$ and simply solve, right?

35. anonymous

where u = 3x ? :)

36. anonymous

right? .-.

37. anonymous

isn't it simpler that way, lol?

38. amistre64

$\int\limits_{} \frac{5(3)}{5(3)} \sin^4 (3x) dx$

39. anonymous

why?

40. anonymous

amistre, i think you're doing it wrong.. and i feel bad caz i'm not actually doing it...

41. amistre64

$\frac{1}{15} \int\limits_{} 5(3) \sin^4 (3x) dx$

42. amistre64

-cos^5 (3x)/15 + C

43. anonymous

use the half angle formula to integrate sin or cos with a power you MUST have sin * cos

44. anonymous

no, amistre, that's definitely wrong T-T

45. anonymous

Why don't you make a substitution of $u=3x \rightarrow dx = \frac{du}{3}$so that then$I=\frac{1}{3}\int\limits_{}{}\sin^4 u du$and now use a reduction formula?

46. amistre64

derive it back again, its right.... it should be right lol

47. anonymous

using the way I cut it down, you'll get : $= \int\limits1 du - 2\int\limits \cos^2u du + \int\limits \cos^2u du$ then you'll use for cos^2 u = 1/2(1+cos2u)

48. anonymous

star is right

49. anonymous

^_^ there, I hope it's right

50. anonymous

as simple as that :)

51. anonymous

sstar is completely right though i have a question where's the person who ASKED this question

52. anonymous

LOL! watching us, hey oneprince did you understand what we did? ^_^

53. anonymous

wait...he left ._.

54. anonymous

yeah -_- but you got it sstar i have integrating cosine sins T-T so messy...

55. anonymous

$I=\frac{1}{3}(\frac{1}{32} (12 u-8 \sin(2 u)+\sin(4 u))+c)$where u = 3x.

56. anonymous

o_o....no wait O_O!

57. anonymous

that's the final answer?

58. anonymous

Sub. u=3x, then du = 3dx --> dx = du/3. Then you have 1/3*int[(sin^4(u) du]

59. anonymous

You can use a reduction formula on sin^4(u)...

60. anonymous

no need for reduction formula ^_^ + i got you lol :)

61. amistre64

what happens when we derive:$- \frac{\cos^5 (3x)}{15} + C$ ??

62. anonymous

why do you want to derive it? that's not the answer lol? ._.

63. anonymous

He wants to see if it matches the integrand.

64. anonymous

oh

65. amistre64

lol.... if its not the answer, then what does it derive to?

66. anonymous

there's only one way to find out, T.R.Y ^_^

67. anonymous

give me a sec, i'll do it

68. anonymous

I won't even try since my page is lagging BADLY

69. anonymous

anyhow, I'm off to review for my DM test tomorrow, good luck all ^_^

70. anonymous

225 cos(3x)*sin(3x)^4

71. anonymous

Omg, I gave you the answer ^^

72. anonymous

wait i lied! i did * 15 instead of / 15 give me another min :P

73. anonymous

74. anonymous

ohh i see why you put that 15 there it's cos(3x)*sin(3x)^4

75. anonymous

cheers :D

76. anonymous

Aren't you trying to find the integral of sin^4(3x)?

77. anonymous

yeah but amistre is trying to show that he's right but i showed that he's wrong ^^

78. amistre64

lol...not trying to show that im right, just trying to figure out what I did wrong :)

79. anonymous

:x sorry i didn't mean it like that but do you get it now?

80. amistre64

I see where my error cropped up ;)

81. anonymous

lols kewls