Integral of $\log (1+a\cos x)$ from 0 to $\pi$

[utkarshakash]

Homework Statement

Evaluate \displaystyle \int_0^{\pi} \log (1+a\cos x) dx

Homework Equations

The Attempt at a Solution

Using Leibnitz's Rule,
F'(a)=\displaystyle \int_0^{\pi} \dfrac{\cos x}{1+a \cos x} dx

Now, If I assume sinx=t, then the above integral changes to
\displaystyle \int_0^{0} \dfrac{dt}{1+a \sqrt{1-t^2}}

Since both the limits are zero now, shouldn't the value of integral be 0!

 

[haruspex]

\displaystyle \int_0^{0} \dfrac{dt}{1+a \sqrt{1-t^2}}

Since both the limits are zero now, shouldn't the value of integral be 0!

No. For one thing, the use of the square root function hides the fact that cos(t) will change sign over the range. Split it into two integrals to be safe.