Leibnitz rule

1. Oct 11, 2012

msmmpts

1. The problem statement, all variables and given/known data

solve ∫log(1+acosx) by differentiation under integral sign (limits are 0 to ∏)

2. Relevant equations

3. The attempt at a solution
=∫(1/1+acosx)cosxdx(by leibinitz by differentiating partially WRT a.

Then how do I proceed,can any one show me all the steps of reduction to standard integral forms??

2. Oct 11, 2012

Simon Bridge

Welcome to PF; :)
what is the base of the logarithm?
by "acos(x)" do you mean $a\cos(x)$ or $\arccos(x)$?
No - because that would be against the rules. But we can try pointing you in promising looking directions in the hope you'll figure it out...

I'd have been tempted to sub:
$e^u=1+\arccos(x) \Rightarrow x=\cos(1-e^u);$
... gets rid of both pesky functions at once.
Then I'd look for a further substitution or explore integrating by parts a couple of times.

The other way ... $e^u=1+a\cos(x)$ ;)

hmmm... actually, Leibnitz's rule looks simpler... back in a tick.

Last edited: Oct 11, 2012
3. Oct 11, 2012

Simon Bridge

$$u(\alpha)=\int_0^\pi \ln|1+\alpha\cos(x)|dx = \int_0^\pi f(x,\alpha)dx= \int_0^\pi \frac{\partial}{\partial\alpha}f(x,\alpha)dx$$ ... note that $\alpha\cos(\pi)=-\alpha$ may be less than -1 making the argument of the logarithm negative.

$$\frac{\partial}{\partial\alpha}f(x,\alpha) =\frac{\partial}{\partial\alpha}\ln|1+\alpha\cos(x)|=\frac{\cos(x)}{1+\alpha\cos(x)}$$... but would this have to be changed if $|\alpha| > 1$

so you end up with: $$u(\alpha)=\int_0^\pi \frac{\cos(x)}{1+\alpha\cos(x)}dx$$... which is where you are at.

well I suppose you could get rid of the trig functions with $z=\cos(x)$ and the identity $\sin(\arccos(z))=\sqrt{1-z^2}$ but I don't hold out much hope.

the other thing to try is $\cos(x)=\frac{1}{2}(e^{ix}+e^{-ix})$

4. Oct 11, 2012

Simon Bridge

I suppose there is always looking for p(x) and q(x) satisfying: $$q^2=1+\alpha\cos(x)$$ and $$q\frac{dp}{dx}-p\frac{dq}{dx}=\cos(x)$$... which will give you a first-order DE in p(x) to solve.

Then, by the quotient rule: $$u(\alpha)=\frac{p}{q}$$... nothing looks tidy and a quick squiz at Wolfram's calculator suggests nothing will be.

[note that $1+\alpha\cos(x)$ can be understood as $|\vec{u}+\vec{v}|$ where $x$ is the angle between the two vectors.]

Last edited: Oct 11, 2012
5. Oct 11, 2012

msmmpts

it is "acosx"..a is the parameter

6. Oct 11, 2012

msmmpts

If I could get rid of cosx from numerator,then
we could have substituted tan(x/2)=t
dx=2dt/1+tsquare
cosx=1-tsquare/1+tsquare.

I tried dividing num&denom by cosx but no good.

7. Oct 11, 2012

Simon Bridge

Ah - Weierstrass substitution is probably better earlier than that:$$\ln|1+\alpha\cos(x)| = \ln|1+t^2 -\alpha(1-t^2)|-\ln|1+t^2|$$... but I suspect we'll end up with the same trouble when we try to find f(t): dx=f(t)dt
since the substitution is $\tan(\frac{x}{2})=t$ then $$dx = \frac{2}{1+t^2}dt$$... still not optimistic [/edit]

I had another look at the substitution: $e^z=1+\alpha\cos(x) \Rightarrow x=-\arccos(\frac{1-e^z}{\alpha})$ because $$dx = -\frac{1}{\alpha}\sqrt{\frac{e^z}{2-e^z}}dz$$... or something.
This would suggest an integration by parts to get rid of the stray z in the integrand, which may lead to hyperbolic functions.

I think that's about the best I can do.
The end result is very messy - the best you can do is try to ease the pain in getting there.

Aside: looks unusually painful for a homework exercise - perhaps there is something in the context that will simplify the problem for you?

Last edited: Oct 11, 2012
8. Oct 11, 2012

msmmpts

simon,according to the exercise the final result does not contain any hyperbolic terms.