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

United States16368 Posts

August 13 2009 05:14 GMT

Did you click on "Show steps", it tells you exactly how one'd go about solving it "by hand"

micronesia

United States24495 Posts

August 13 2009 05:15 GMT

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?

Peter[Deuce]

United States93 Posts

August 13 2009 05:19 GMT

I'm thinking u substitution but don't know how to manipulate the function to use it.

hype[NZ]

Japan412 Posts

August 13 2009 05:37 GMT

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]

United States93 Posts

August 13 2009 05:39 GMT

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

United States121 Posts

August 13 2009 05:59 GMT

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

blue_arrow

1971 Posts

August 13 2009 06:20 GMT

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

coltrane

Chile988 Posts

August 13 2009 06:22 GMT

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

caldo149

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.=)

Ichigo1234551

United States649 Posts

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

igotmyown

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

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

United States253 Posts

August 13 2009 08:03 GMT

#14

for tans / cotans i always put in terms of cos and sin

helps me.

Hiphopapotamus

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