Solving an integral with parameter: $\int_{0}^{\pi/2} \frac{\ln(1+a\cos x)}{\cos x}dx$

Question

I have this integral with parameter:

$$I(a) =\int_{0}^{\pi/2} \frac{\ln(1+a\cos x)}{\cos x}dx, 0

Tried to use the differentation under the integral sign:

$$\frac{\partial I}{\partial a} = \frac{1}{a\cos x+1}$$

$$\int_{0}^{\pi/2}\frac{dx}{a\cos x+1} = \frac{2\tanh^{-1}\left(\frac{(-1 + a) \tan(x/2)}{\sqrt{a^2-1}}\right)}{\sqrt{a^2-1}} + C$$

I think that something goes wrong on this step. If I substitute ${a}$ into $\sqrt{a^2-1}$, the result is negative.

Any help would be really helpful.

Answer 1

Your work so far is completely fine; you just forgot to apply the definite integral.
Note that $\tanh(x/i) = \tan(x)/i$, so that $\tanh^{-1}(x/i) = \tan^{-1}(x)/i$.

We now have $$ \frac{\partial I}{\partial a} = \left. \frac{2\tanh^{-1}\left(\frac{(-1 + a) \tan(x/2)}{\sqrt{a^2-1}}\right)}{\sqrt{a^2-1}} \right|_{x=0}^{x = \pi/2} = \frac{2\tanh^{-1}\left(\frac{a-1}{\sqrt{a^2-1}}\right)}{\sqrt{a^2-1}}=\\ \frac{2\tanh^{-1}\left(\frac{a-1}{i\sqrt{1-a^2}}\right)}{i\sqrt{1-a^2}} = \frac{2\tan^{-1}\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{i^2\sqrt{1-a^2}} = \\ -\frac{2\tan^{-1}\left(\frac{a-1}{\sqrt{1-a^2}}\right)}{\sqrt{1-a^2}} $$ And clearly, $I|_{a = 0} = 0$. From there, integrate with substitution $u = \sqrt{\frac{1-a}{1+a}}$.