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[Help] Calculus:Integration

Blogs > Peter[Deuce]
Post a Reply
Peter[Deuce]
Profile Blog Joined November 2007
United States93 Posts
Last Edited: 2009-08-13 05:12:46
August 13 2009 05:11 GMT
#1
This problem is going to be on my final and I need every point I can get to get an A =(

http://img205.imageshack.us/img205/7389/integrate.jpg

∫(1+4cot(x))/(4-cot(x)) dx from pi/4 to pi/2

Some guy on yahoo answers solved it by going to
http://www.wolframalpha.com/input/?i=∫(1+4cot(x))/(4-cot(x)) dx
but I'm pretty sure you can solve it by hand.

Edit: I did go to the wolframalpha site and looked up the "Show Steps" but it seemed far too complicated to actually be the logical way and human way to solve it.

*
Cambium
Profile Blog Joined June 2004
United States16368 Posts
August 13 2009 05:14 GMT
#2
Did you click on "Show steps", it tells you exactly how one'd go about solving it "by hand"
When you want something, all the universe conspires in helping you to achieve it.
micronesia
Profile Blog Joined July 2006
United States24645 Posts
August 13 2009 05:15 GMT
#3
Is it possible your textbook teaches you how to solve this intergral? Could you at least make some guesses as to what techniques or formulas you might use?
ModeratorThere are animal crackers for people and there are people crackers for animals.
Peter[Deuce]
Profile Blog Joined November 2007
United States93 Posts
August 13 2009 05:19 GMT
#4
I'm thinking u substitution but don't know how to manipulate the function to use it.
hype[NZ]
Profile Blog Joined September 2008
Japan412 Posts
August 13 2009 05:37 GMT
#5
Yep you are right that you will need to do a substitution, however you obviously need to decompose the integrand into something where it clear that the substitution will be useful.

I did do this problem just now, and it's not too much work. One thing to try when you are stuck on these sorts of problems is to decompose the cotangents etc into their basic cosine and sine components, rearrange a bit and then you might see something useful

Good luck!
Peter[Deuce]
Profile Blog Joined November 2007
United States93 Posts
August 13 2009 05:39 GMT
#6
Ah what a simple solution by just multiplying both sides by sinx. I kept thinking it was much more complicated. I was even considering integrating it by parts. thanks
Hiphopapotamus
Profile Joined July 2009
United States121 Posts
August 13 2009 05:59 GMT
#7
Teachers love making a simple solution to some of the more ugly looking problems. On a side note, how did you know this is going to be on your final? hax
My lyrics are bottomless!
blue_arrow
Profile Blog Joined July 2008
1971 Posts
August 13 2009 06:20 GMT
#8
plenty of teachers will tell you questions that are gonna be on the final

the point is to force everyone to learn the answer and learn about the simplicity and beauty of it all

besides, any good teacher doesn't really give a damn about the marks
| MLIA | the weather sucks dick here
coltrane
Profile Blog Joined June 2008
Chile988 Posts
August 13 2009 06:22 GMT
#9
You know this?

(u*v)` = u`*v + u*v`


∫(u*v)`dx = ∫(u`*v)dx + ∫(u*v`)dx

u*v = ∫(u`*v)dx + ∫(u*v`)dx

∫(u`*v)dx = u*v - ∫(u*v`)dx
Jävla skit
caldo149
Profile Blog Joined April 2009
United States469 Posts
August 13 2009 07:02 GMT
#10
from a first glance, i suggest trying to split the fraction into the sum/difference of 2 fractions, and/or multiplying by the the conjugate on top and bottom (so technically you multiply by 1 and it's legal)
though not necessarily in that order. Then check for pythagorean identity...?

if you don't know: conjugate of 1+4cotx is 1-4cotx, conjugate of 4-cotx is 4+cotx

I'm too lazy to actually try this right now... but i hope this helped. hfgl on the final.=)
Hellions are my homeboys
Ichigo1234551
Profile Blog Joined October 2008
United States649 Posts
Last Edited: 2009-08-13 07:53:09
August 13 2009 07:47 GMT
#11
okay ima try to solve this, but i havnt touched my math stuff in awhile

damn it i took 2 months of calculus and i forgot everything
I WILL DESTROY YOU IN 2009 OK???????????????
igotmyown
Profile Blog Joined April 2009
United States4291 Posts
August 13 2009 07:50 GMT
#12
cot(x)=y
1/sin(x)^2 dx = dy
y^2+1=1/sin(x)^2

dx=dy*sin(x)^2
dx=dy*1/(1+y^2)

∫(1+4cot(x))/(4-cot(x)) dx from pi/4 to pi/2
∫(1+4y)/(4-y)*dy/(1+y^2) from 1 to 0 (it's backwards)
This can be solved via logarithms with 4-y, 1+yi, 1-yi, I forget what the method's called

Or you can use clever trig substitutions
igotmyown
Profile Blog Joined April 2009
United States4291 Posts
August 13 2009 07:56 GMT
#13
tan (x/2) = sin x/(1 + cos x)
∫(1+4cot(x))/(4-cot(x)) dx from pi/4 to pi/2
1+4*cos/(1+sin)/(4-cos/(1+sin))
(1+sin)+4*cos/(4+4sin-cos)

cot(a)=1/4
∫(cot(a)+cot(x))/(1-cot(a)*cot(x)) dx from pi/4 to pi/2
∫(cot(a+x) dx from pi/4 to pi/2, where a=cot^(-1)(1/4)
mcgriddle
Profile Joined April 2009
United States253 Posts
August 13 2009 08:03 GMT
#14
for tans / cotans i always put in terms of cos and sin helps me.
Reason obeys itself....and ignorance submits to whatever is dictated to it.
Hiphopapotamus
Profile Joined July 2009
United States121 Posts
August 13 2009 08:19 GMT
#15
I think he already solved it. I did:

Multiply top and bottom by sinx and let
u = 4sinx - cosx
du = sinx + 4cosx dx

-> ∫ du/u
-> plug into log for answer
My lyrics are bottomless!
Mah Buckit!
Profile Joined April 2009
Finland474 Posts
August 13 2009 09:21 GMT
#16
On August 13 2009 17:03 mcgriddle wrote:
for tans / cotans i always put in terms of cos and sin helps me.


My math teachers always said that cotans are bloody useless and so we never used them

Starcraft? Epic Grimness.
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