Question: Calculate $I(y) = \int_0^{\pi} \ln(1+y\cos x)dx$.
I found this problem in my professor's calculus document and tried to do it many times but still felt uncertain about my answer
Here my attempt:
Let $f(x,y) = ln(1+y\cos x)$ which is continuous on $[0,\pi] \times (0,1]$, we have: $f'_{y} = \frac{\cos x}{1+y\cos x}$ still continous on $[0,\pi] \times (0,1)$; therefore, $I(y)$ is continuous on $(0,1]$ and differentiable on $(0,1)$
Thus, we have: $$I'(y) = \int_0^\pi f'_{y}(x,y) dx = \int_0^\pi\frac{\cos x}{1+y\cos x} dx $$ Considering: $$J = \int \frac{\cos x}{1+ y \cos x}dx = \int \frac{\cos x \over y}{{1\over y}+\cos x}dx = \int \frac{a\cos x}{a + \cos x}dx\ \left(let\ a={1\over y}\right)\\ = a\int\left(1-\frac{a}{a + \cos x} \right)dx\\=ax - a^2\int \frac{1}{a + \cos x}dx $$ Here $$\int \frac{1}{a + \cos x}dx = \int \frac{\frac{2dt}{1+t^2}}{a + \frac{1-t^2}{1+t^2}}\ \left(let\ t = \tan {x\over2}\ and\ using\ Weierstrass\ substitution\right)\\ = \int \frac {2}{a(1+t^2) + 1 - t^2}dt = \int \frac{2}{(a-1)t^2+a+1}dt = {2\over{a-1}}\int \frac{1}{t^2 + \frac{a+1}{a-1}}dt\\={2\over{a-1}}\cdot{{\sqrt{a-1}}\over{\sqrt{a+1}}}\cdot\arctan\left(\frac{t\cdot\sqrt{a-1}}{\sqrt{a+1}}\right) = {2\over{\sqrt{a^2-1}}}\cdot \arctan\left(\frac{\tan {x\over2}\cdot\sqrt{a-1}}{\sqrt{a+1}}\right)$$
Therefore$$J = ax - {2a^2\over{\sqrt{a^2-1}}}\cdot \arctan\left(\frac{\tan {x\over2}\cdot\sqrt{a-1}}{\sqrt{a+1}}\right)\\ =\frac{x}{y} - {2\over{y\sqrt{1-y^2}}}\cdot \arctan\left(\frac{\tan {x\over2}\cdot\sqrt{1-y}}{\sqrt{1+y}}\right) $$ Thus$$I'(y) = \left.{\left(\frac{x}{y} - {2\over{y\sqrt{1-y^2}}}\cdot \arctan\left(\frac{\tan {x\over2}\cdot\sqrt{1-y}}{\sqrt{1+y}}\right)\right)}\right\rvert_0^\pi\\=\frac{\pi}{y}-{2\over{y\sqrt{1-y^2}}}\cdot \lim_{x \to \pi^{-}}\arctan\left(\frac{\tan {x\over2}\cdot\sqrt{1-y}}{\sqrt{1+y}}\right)=\frac{\pi}{y}-{2\over{y\sqrt{1-y^2}}}\cdot \frac{-\pi}{2}\\ = \frac{\pi}{y}+{\pi\over{y\sqrt{1-y^2}}}\\ \Rightarrow I(y) = \int \left(\frac{\pi}{y}+{\pi\over{y\sqrt{1-y^2}}}\right)dy = \pi \ln(y)+\pi \ln\left[\tan \left(\frac{\arcsin y}{2}\right)\right]+C$$ Because $I(y)$ is continuous on $(0,1]$, we have: $$I(1) = \int_0^\pi \ln(1+\cos x)dx = 0 + 0 + C = -\pi\ln(2)\ \left(\int_0^\pi \ln(1+\cos x)dx = -\pi\ln(2)\right) $$ In conclusion
$$I(y) = \pi \ln(y)+\pi \ln\left[\tan \left(\frac{\arcsin y}{2}\right)\right]-\pi\ln(2) \\=\pi\ln\left[\frac{y}{2}\tan \left(\frac{\arcsin y}{2}\right)\right]$$
However. I'm not sure whether the interval I choose for the function is true or not, so that I could apply for the latter calculation. Please give it a look and give a thought, Thank you!
