anonymous one year ago solve this integral :/ not sure how to do this!

1. anonymous

$\int\limits_{} \sin^5 t \cos^4 t dt$

2. Zale101

|dw:1442984817219:dw|

3. Astrophysics

Some general rules with these trig functions |dw:1442984972414:dw|

4. Zale101

Any ideas of what to do next?

5. Zale101

That chart is very helpful! @astrophysics

6. Astrophysics

Yeah, it's pretty awesome :P

7. Astrophysics

http://puu.sh/kl22A/732d78a86f.png if anyone wants the link XD

8. anonymous

sorry!! i am back!! :) i am reading the chart right now :) one sec!! hopefully i can have a clue what happens next:)

9. anonymous

okay, i see now what you did @Zale101 so i strip one sin away and get|dw:1442985650253:dw|

10. anonymous

not sure what to do next though :/

11. Astrophysics

|dw:1442985664022:dw|

12. anonymous

ohh so i would get this? |dw:1442985687508:dw|

13. Zale101

|dw:1442985746191:dw|

14. anonymous

okay!! i forget though, from here, do i sub something in? :/

15. Zale101

U-Sub! :)

16. anonymous

so we get integral sin t (sin^2 t)^2 u^4 dt ?

17. Zale101

Nope. First of all, you cant have u while the rest of the terms are in t. Second, you have to use trig identities for (sin^2t)^2

18. anonymous

ohhh how do i do that then?

19. Zale101

$$(sin^2t)^2=(1-cos^2t)^2$$

20. anonymous

ohh the square thing?

21. Zale101

$$sin^4t=(sin^2t)^2=(1−cos^2t)^2$$

22. anonymous

confused which trig functions to insert in now?

23. Zale101

Yes. If you wan to replace sin^2 with the trig identities 1-cos^2t. But sin is squared, so the identity has to be squared as well.

24. anonymous

ohh but it is fine leaving like sin? what happens next?

25. Zale101

Next, you do u-sub. But first show me your integral now after using trig identities.

26. anonymous

$\int\limits_{} (1-\cos^2 t)((1-\cos^2)^2 t)^2 cost^4 t dt$ this?

27. Zale101

$\int\limits_{}^{}\sin(t) (\sin^2(t))^2\cos^4(t)dt$ $\int\limits_{}^{}\sin(t) (1-cos^2(t))^2\cos^4(t)dt$

28. anonymous

ohh okay oopsies :P what happens now? :O

29. Zale101

What you did was wrong because you can replace sint with 1-cos^2 trig identities say $$sin^2t=1-cos^2$$

30. anonymous

ohhhh:O

31. Zale101

$$sin^2t+cos^2t=1$$

32. Zale101

What is your next step now?

33. anonymous

i integrate? :/

34. Zale101

How?

35. anonymous

i am not sure :(

36. anonymous

i am sorry :( i am very bad at calc:(

37. Zale101

We need u-sub here. What would be you u in this case?

38. Zale101

your*

39. anonymous

u=sin(t) ?

40. Zale101

Would u=sint work?

41. anonymous

i think so :/

42. Zale101

I will try u=sint and see if it works.

43. anonymous

ok!! sorry, i get very confused with these concepts :/

44. Zale101

|dw:1442986993486:dw|

45. anonymous

ohhh okay :/ what should we do then?

46. Zale101

try u=cos(t)

47. Astrophysics

A saying I remember, get the odd man out! You want the odd man out!

48. anonymous

okay!! where do we sub that into ? :/

49. Zale101

With integration, here is what you should do. Every time you make an educated guess on what might the u be, do the integration and see if it works.If not, then try something else. This is what i did with integral and it was helpful.

50. Zale101

integral problems*

51. anonymous

ohhh okie :) so we should use u=cos(t) and so we do this? $\int\limits_{}sint(1-u)^2 u^4 dt$

52. Zale101

You cant place in the u when you have not finished u-subbing

53. anonymous

ohh no! I'm sorry, i don't know what to do here :(

54. Astrophysics

When you do a u sub, you must also get the derivative, that's what will cancel out the sint $u= cost \implies du= - sint dt$

55. Astrophysics

You don't want t and u's, just one variable

56. anonymous

okay! what would it look like then?

57. Zale101

Do u=cost and du=-sint dt and tell me what you will get

58. anonymous

i get this? is it correct? integral with plug in u and du udu and i sub in and replace and get this? -1/5cos^5t+2/7cost^7t-1/9cos^9t+C?

59. anonymous

thank you so much @Zale101 @Astrophysics i will try to figure this one out some more tomorrow! i need to sleep now because it is 2am here and i have to be up in four and a half hrs!! thanks so much for your help!!! :D sorry it was hard for me to understand!! have a nice night!!

60. Zale101

Alright, good luck and take care!

61. Astrophysics

Good night! :)

62. Zale101

Just a type error i made "What you did was wrong because you $$\color{red}{can't}$$ replace sint with 1-cos^2"