Question

I was working on my practice final and came up with a solution to one of the problems. My math teacher just released the answers yesterday and he used a different method than I did and got a different answer. I have been unable to reach him and my final is tomorrow morning. Could someone please check my answer for the question below.

Answer 1

\[\int\limits_{}^{}\frac{ cosx +1}{ sinx +2}\]

Answer 2

just integrate?

Answer 3

I forgot the dx piece but that is the question I'm typing my solution now. Yes just integrate.

Answer 4

use u-substitution: u=sin(x)+2 where du=cos(x)dx
We can then separate the numerator with the same denominator

Answer 5

I started by separating the integral in two like so
\[\int\limits_{}^{}\frac{ cosx }{ sinx+2 }dx + \int\limits_{}^{}\frac{ 1 }{ sinx+2 }\]

Answer 6

yep that's right. then use u-substitution from what i mentioned

Answer 7

yes and then I used the trig substitution on the RHS integral of tan(x/2)=t

Answer 8

Which means
\[sinx=\frac{ 2t }{ t^2 +1 }\]
and
\[dt=\frac{ 2 }{ t^2+1 }\]

Answer 9

ok, then what did you get?

Answer 10

so we have
\[\int\limits_{}^{} \frac{1}{ \frac{ 2t }{ t^2+1 }+2 }\times \frac{ 2 }{ t^2+1}dx\]

Answer 11

it's getting messier, but go on :)

Answer 12

Compared to his method . . . This is quite clean. Mine takes a quarter of a page his takes close to two

Answer 13

wanna know how i'll do this?

Answer 14

I simplified bring the denominator of the LHS fraction to a common denominator and bringing up that common denominator to get
\[\int\limits_{}^{}\frac{ t^2+1 }{ 2(t^2+t+1) }\times \frac{ 2 }{ t^2+1 }\]

Answer 15

sure if its easier then my method . . . just remember that the +1 in the numerator of the original question makes it very difficult to do a different trig substitution in my opinion

Answer 16

but but it is about to get a whole lot nicer as I am sure you can see now.

Answer 17

It comes down to this because the other integral was trivial.\[
\int\limits_{}^{}\frac{ 1 }{ \sin (x)+2 }dx
\]

Answer 18

yes my teacher did the trig substitution before separating the integral and it made for a very long final answer.

Answer 19

I can post his final answer if you like.

Answer 20

Well, \[
\int\limits_{}^{}\frac{ 1 }{ \sin (x)+2 }dx = \int\limits_{}^{}\frac{ \sin^2(x)+\cos^2(x) }{ \sin (x)+2 }dx
\]Then doing \(u=\sin(x)+2\)...

Answer 21

hmmm it might help a bit.

Answer 22

I think I may have tried this method and got back to the integral I was about to get when I finished cancelling above along with another another integral. I could be wrong though I probably used close to dozen sheets of paper and did not save all of them.

Answer 23

I am wondering now if separating them was a good idea...

Answer 24

wait no I did try this method but ran into the problem of pulling out a du. If you have an easier/faster solution without separating them I'm all for it. I only have 2 hours to write the test so the quicker I can get a difficult problem the better.

Answer 25

I started off by seperating them just to isolate the part that was giving me trouble with computing the integral (the +1 in the numerator).

Answer 26

oops sorry * separating

Answer 27

\[\int\limits_{}^{}\frac{ \cos(x)+1 }{ \sin(x)+2 }dx\]u=sin(x)+2 ---> du=cos(x)dx
\[\int\limits_{}^{}\frac{ \cos(x)+1 }{ u }\frac{ du }{ \cos(x) }\]\[\int\limits_{}^{}\frac{ \cos(x) }{ u }\frac{ du }{ \cos(x) }+\int\limits_{}^{}\frac{ 1 }{ u }\frac{ du }{ \cos(x) }\]\[\int\limits\limits_{}^{}\frac{ \cos(x) }{ ucos(x) }du \rightarrow \int\limits\limits_{}^{}\frac{ 1 }{ u }du \rightarrow \ln(u)+(someconstant)\]
lol, now i'm confused

Answer 28

Actually my math 115 teacher just sent me and email about five minutes ago saying my method was correct and I did not have an error. So I guess this is now a find the fastest way to do the question now but I should be fine.

Answer 29

@Lynncake My math teacher always tells us to get rid of every one of the old variable if we are substituting I think his reason for telling us this is the confusion you are feeling now.

Answer 30

yeah -_- that's why i hate integrating with trig identities in it O_O @whatisthequestion