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9/22/2017 12:11:25 AM
Posted: 3/5/2006 1:59:02 PM EDT
[Last Edit: 3/5/2006 4:01:13 PM EDT by JL7]
can anybody help me with this or tell me if my answer is correct

Link Posted: 3/5/2006 2:09:09 PM EDT
87 ?
Link Posted: 3/5/2006 2:11:33 PM EDT
Nope, That's incorrect. You need to use Trigonometric substution for that integral.
Link Posted: 3/5/2006 2:12:05 PM EDT
Doing calculus isn't like riding a bike. You do forget if you don't use it. Sorry, I might have been able to help you 10 years ago when I was in college, but today I'm lost.

Do you have access to something like matlab? I bet it could help.
Link Posted: 3/5/2006 2:14:03 PM EDT
i did i sepperated into cos^2sin^2sinx and then used sin^2+cos^2 = 1, then u substitution
u = cosx
Link Posted: 3/5/2006 2:16:53 PM EDT
Sorry, once we studied Pi, my life took a hole new direction...
Link Posted: 3/5/2006 2:17:16 PM EDT
[Last Edit: 3/5/2006 2:17:56 PM EDT by Red_Beard]
Too hard for me.



Why don't you try taking the derivative of the right side of your equation?
Link Posted: 3/5/2006 2:19:21 PM EDT
Your method and answer are correct.
Link Posted: 3/5/2006 2:19:24 PM EDT
I thought this was a math question, all I see is GREEK.
Link Posted: 3/5/2006 2:20:11 PM EDT
[Last Edit: 3/5/2006 2:22:30 PM EDT by krpind]

Originally Posted By TheRegulator:
87 ?




Originally Posted By DrCaligari1:
Nope, That's incorrect. You need to use Trigonometric substution for that integral.






ETA..DrCaligari1 I'm pretty sure you weren't replying to TheRegulator, but it fit so perfectly, I couldn't resist.


Link Posted: 3/5/2006 2:22:23 PM EDT
The correct answer appears to be: 1/70 cos^5(x)(5cos(2x)-9)

Don't ask me why!


The Integrator
Link Posted: 3/5/2006 2:22:48 PM EDT

Originally Posted By JL7:
i did i sepperated into cos^2sin^2sinx and then used sin^2+cos^2 = 1, then u substitution
u = cosx



Yes, u use u substitution knowing the relation of cos and sin.
Link Posted: 3/5/2006 2:23:02 PM EDT
Hi,

For integrals in the form of sin(x)^a*cos(x)^b where the power of cos is odd, use cos(x)^2=1-sin(x)^2 . This integral will require several substitutions and rearrangements before it will be solveable by a u substution of u=cos(x) or u=sin(x). I'd work the problem out for you but ar15 has no Integral symbol 6
Link Posted: 3/5/2006 2:23:03 PM EDT

Originally Posted By fssf158:
Your method and answer are correct.



did you work the problem anyway thanks i hope your right


Link Posted: 3/5/2006 2:26:23 PM EDT

Originally Posted By DrCaligari1:
Hi,

For integrals in the form of sin(x)^a*cos(x)^b where the power of cos is odd, use cos(x)^2=1-sin(x)^2 . This integral will require several substitutions and rearrangements before it will be solveable by a u substution of u=cos(x) or u=sin(x). I'd work the problem out for you but ar15 has no Integral symbol .

6



you could use {
Link Posted: 3/5/2006 2:27:14 PM EDT
Yes, I worked the problem. I also checked my calc. book and found a virtually identical example (odd power of sine, even power of cosine).
Link Posted: 3/5/2006 2:37:08 PM EDT

Originally Posted By fssf158:
Yes, I worked the problem. I also checked my calc. book and found a virtually identical example (odd power of sine, even power of cosine).




Yep, He's right. Just worked the problem. I can post the worked out solution if you want.


Link Posted: 3/5/2006 2:39:36 PM EDT
no need if i got the right answer. thank you both

Link Posted: 3/5/2006 2:54:22 PM EDT
If you have a TI-83, or something similar, just graph your answer and the derivative of your answer (or the other way around). If they seem to be the same function, you're probably right. It's not particularly rigorous, but it works in the general case.
Link Posted: 3/5/2006 2:55:47 PM EDT


This kind of language will not be tolerated here!!!!
Link Posted: 3/5/2006 2:58:27 PM EDT

Originally Posted By AZ-K9:
I thought this was a math question, all I see is GREEK.



My preference is egyptian - all the pretty pictures of animals and such.
Link Posted: 3/5/2006 4:00:41 PM EDT
so this is right before we start with partial fractions, any ideas which one i can do without partial fractions, and how about to get started

Link Posted: 3/5/2006 4:07:58 PM EDT
btt
Link Posted: 3/5/2006 4:10:24 PM EDT

Originally Posted By bookertbab:
Doing calculus isn't like riding a bike. You do forget if you don't use it. Sorry, I might have been able to help you 10 years ago when I was in college, but today I'm lost.

Do you have access to something like matlab? I bet it could help.



+1


Especially if it was Calc II

Differential Equations was easier for me
Link Posted: 3/5/2006 4:14:42 PM EDT
it is calc 2

matlab and mathcad both, but they dont show you how to work the problem. And they dont always give the answer im looking for. Kinda like the TI-89
Link Posted: 3/5/2006 4:15:24 PM EDT
[Last Edit: 3/5/2006 4:16:19 PM EDT by Yossarian]
WTF man....you need a linguist......THATS GREEK!!!!!



Errrr: nice AZ-K9
Link Posted: 3/5/2006 4:31:27 PM EDT
anyone
Link Posted: 3/5/2006 4:38:40 PM EDT
[Last Edit: 3/5/2006 4:48:14 PM EDT by DrCaligari1]

Originally Posted By JL7:
so this is right before we start with partial fractions, any ideas which one i can do without partial fractions, and how about to get startedhref=img378.imageshack.us/img378/843/math22zf.jpg





I would solve both of these with partial fractions. (A) requries partial fractions, but (B) can be solved by logirithmic integration, where u=x^2-6x+5 and du=2x-6 . The derivative of ln(u) is 1/u and the integral of du/u is ln(u)

notice in B that the deravitive of the denominator is 2x-6. this only varies from the numerator by a constant of 2. Set up the problem as 1/2* (du/u) and integrate. You will get 1/2ln(u). Subs in u=x^2-6x+5 and you will get 1/2[ln(x^2-6x+5)].

You'll see that when you learn partial fractions that it is a much easier way to solve these.
Link Posted: 3/5/2006 7:47:21 PM EDT
Oh Puhleeze...you can solve this through several iterations of Integration by Parts...
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