attn math geniuses (Update: New Problem)

General » General Discussion

Site Notices

TNVC Partners With Envision Technology - MARS Rangefinder & RAIL Family, MARS-Lc In-Stock!

[ARCHIVED THREAD] - attn math geniuses (Update: New Problem)

Posted: 3/5/2006 1:59:02 PM EDT

can anybody help me with this

or tell me if my answer is correct

Posted: 3/5/2006 2:09:09 PM EDT

[#1]

87 ?

Posted: 3/5/2006 2:11:33 PM EDT

[#2]

Nope, That's incorrect. You need to use Trigonometric substution for that integral.

Posted: 3/5/2006 2:12:05 PM EDT

[#3]

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.

Posted: 3/5/2006 2:14:03 PM EDT

[#4]

i did i sepperated into cos^2sin^2sinx and then used sin^2+cos^2 = 1, then u substitution
u = cosx

Posted: 3/5/2006 2:16:53 PM EDT

[#5]

Sorry, once we studied Pi, my life took a hole new direction...

Posted: 3/5/2006 2:17:16 PM EDT

[#6]

Too hard for me.

Why don't you try taking the derivative of the right side of your equation?

Posted: 3/5/2006 2:19:21 PM EDT

[#7]

Your method and answer are correct.

Posted: 3/5/2006 2:19:24 PM EDT

[#8]

I thought this was a math question, all I see is GREEK.

Posted: 3/5/2006 2:20:11 PM EDT

[#9]

Quoted:
87 ?

Quoted:
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.

Posted: 3/5/2006 2:22:23 PM EDT

[#10]

The correct answer appears to be: 1/70 cos^5(x)(5cos(2x)-9)

Don't ask me why!

The Integrator

Posted: 3/5/2006 2:22:48 PM EDT

[#11]

Quoted:
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.

Posted: 3/5/2006 2:23:02 PM EDT

[#12]

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

Posted: 3/5/2006 2:23:03 PM EDT

[#13]

Quoted:
Your method and answer are correct.

did you work the problem

anyway thanks i hope your right

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

[#14]

Quoted:
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 {

Posted: 3/5/2006 2:27:14 PM EDT

[#15]

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).

Posted: 3/5/2006 2:37:08 PM EDT

[#16]

Quoted:
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.

Posted: 3/5/2006 2:39:36 PM EDT

[#17]

no need if i got the right answer. thank you both

Posted: 3/5/2006 2:54:22 PM EDT

[#18]

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.

Posted: 3/5/2006 2:55:47 PM EDT

[#19]

This kind of language will not be tolerated here!!!!

Posted: 3/5/2006 2:58:27 PM EDT

[#20]

Quoted:
I thought this was a math question, all I see is GREEK.

My preference is egyptian - all the pretty pictures of animals and such.

Posted: 3/5/2006 4:00:41 PM EDT

[#21]

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

Posted: 3/5/2006 4:07:58 PM EDT

[#22]

btt

Posted: 3/5/2006 4:10:24 PM EDT

[#23]

Quoted:
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

Posted: 3/5/2006 4:14:42 PM EDT

[#24]

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

Posted: 3/5/2006 4:15:24 PM EDT

[#25]

WTF man....you need a linguist......THATS GREEK!!!!!

Errrr: nice AZ-K9

Posted: 3/5/2006 4:31:27 PM EDT

[#26]

anyone

Posted: 3/5/2006 4:38:40 PM EDT

[#27]

Quoted:
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.

Posted: 3/5/2006 7:47:21 PM EDT

[#28]

Oh Puhleeze...you can solve this through several iterations of Integration by Parts...

[ARCHIVED THREAD] - attn math geniuses (Update: New Problem)

General » General Discussion

Win a FREE Membership!

Sign up for the ARFCOM weekly newsletter and be entered to win a free ARFCOM membership. One new winner* is announced every week!

You will receive an email every Friday morning featuring the latest chatter from the hottest topics, breaking news surrounding legislation, as well as exclusive deals only available to ARFCOM email subscribers.

Warning

Confirm Action

About AR15.COM

Stay Connected

Newsletter

Contact Us