Question

This is continues

Answer 1

http://openstudy.com/study#/updates/54a30f39e4b0a96030d7a6a0
is the initial post.

Answer 2

I have:
\[\frac{1}{4} \int\limits_{ }^{} \frac{\ln p}{(1-p)\sqrt{p}}~dp\]

Answer 3

(I did several substitutions in my post, and made some typos. that is all for that link)

Answer 4

what do I do now?

Answer 5

Wait, did you say the original problem you had was \[\Large \int\limits i^{\cos x} dx\]

Answer 6

no, I said that it is not anything like silly imaginary integral

Answer 7

my original prob, as in that link, was,
\[\int\limits_{ }^{ } \ln(\cos v)~dv\]

Answer 8

I have to brb but I'll be back soon

Answer 9

sure

Answer 10

Absolutely fascinated!

Answer 11

tnx:)

Answer 12

I wanted to attempt partial fractions, but am I permitted to do so (provided I don't plug in 1 or 0 for p, into A(sqrt{1-p})+B*p=ln(p) ) ??

Answer 13

\[\frac{1}{4} \int\limits_{ }^{ } \frac{\ln p}{(1-p)\sqrt{p}}~dp\]\[\frac{A}{\sqrt{p}}+\frac{B}{1-p}\]\[A(1-p)+B \sqrt{p} =\ln(p)\]

Answer 14

when
p=e,
and
p=e^2
\[A(1-e)+B \sqrt{e} =1\]\[A(1-e^2)+B e=2\]

Answer 15

multiply the first equation times -(1+e)
\[-A(1-e^2)-B(1+e)\sqrt{e}=-e-1\]

Answer 16

what in the world am I... but will see if I can, I don't see any restrictions, asides from difficulty.

Answer 17

adding the 2 equations.
\[B \sqrt{e}-(1+e)B \sqrt{e}=1-e\]

Answer 18

\[B \sqrt{e}-(1+e)B \sqrt{e}=1-e\]\[-e~B \sqrt{e}=1-e\]\[B=\frac{e-1}{e \sqrt{e}}\]

Answer 19

\[A(1-e)+B \sqrt{e}=1 \]
\[A(1-e)+ \frac{e-1}{e }=1\]

Answer 20

\[A(1-e)+ \frac{e-1}{e }=\frac{e}{e}\]\[A(1-e)=\frac{1}{e}\]\[A=\frac{1}{e(1-e)}\]

Answer 21

Does this involve integration by parts?

Answer 22

\[\frac{1}{4} \int\limits_{ }^{ } \frac{\frac{1}{e(1-e)}}{\sqrt{p}}+\frac{\frac{e-1}{e \sqrt{e}}}{1-p}~dp\]

Answer 23

no, thi is partial fractions

Answer 24

thx

Answer 25

and then I get,
\[\frac{2\sqrt{p}}{4e(1-e)}+\frac{(e-1)~\ln (1-p)}{4e \sqrt{e}}+C\]

Answer 26

yes, got it now

Answer 27

and then sub in what I had before p=cos^2v.

Answer 28

\[\frac{2 \cos v}{4e(1-e)}+\frac{(e-1)~\ln (\sin^2v)}{4e \sqrt{e}}+C\]

Answer 29

Jesus your good fbi2015!

Answer 30

I am not sure if what I did is correct though

Answer 31

tnx for the complement though.

Answer 32

I am not done

Answer 33

\[\frac{2 \cos v}{4e(1-e)}+\frac{(e-1)~\ln (\sin v)}{2e \sqrt{e}}+C\]

Answer 34

this is what I am arriving at.

Answer 35

solved my BS prob, closing.