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

I have this integral with parameter:

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

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.

• how can the result of your final integral depend on $x$? Commented Feb 9, 2017 at 11:07
• You have the right idea; I think you just made a mistake integrating $\int \frac{dx}{a \cos(x) + 1}$ Commented Feb 9, 2017 at 11:10
• Actually, your work so far is completely fine. Commented Feb 9, 2017 at 11:18

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 $u = \sqrt{\frac{1-a}{1+a}}$.

• is the next integration so obvious? furthermore i think the sign of your squareroots is somehow of ...at least it is not obvious to a beginner how to make sense out of $\sqrt{a^2-1}$ for $0<a<1$ Commented Feb 9, 2017 at 11:25
• see my improved comment Commented Feb 9, 2017 at 11:28
• You're right, it's not obvious Commented Feb 9, 2017 at 11:28
• to tackle the integration w.r.t to $a$, try $\sqrt{1-a}/\sqrt{1+a}=q$ the result is very pleasing Commented Feb 9, 2017 at 11:30
• See my latest edit. Commented Feb 9, 2017 at 11:36

Let’s use Feynman’s Technique of Integration by defining $$I(a):=\int_{0}^{\frac{\pi}{2}} \frac{\ln (1+a \cos x)}{\cos x} d x$$

where $$0\leq a \leq 1.$$ Differentiating $$I(a)$$ w.r.t. $$a$$ yields $$I^{\prime}(a)=\int_{0}^{\frac{\pi}{2}} \frac{1}{1+a \cos x} d x$$ Letting $$t=\tan \dfrac{x}{2}$$, then \begin{aligned} I^{\prime}(a) &=2 \int_{0}^{1} \frac{d t}{1+a+(1-a) t^{2}} \\ &=\frac{2}{1-a} \int_{0}^{1} \frac{d t}{t^{2}+(\sqrt{1+a})^{2}} \\ &=\frac{2}{1-a} \cdot \sqrt{\frac{1-a}{1+a}}\left[\tan ^{-1}\left(t \sqrt{\frac{1-a}{1+a}}\right)\right]_{0}^{1}\\ &=\frac{2}{\sqrt{1-a^{2}}} \tan ^{-1}\left(\sqrt{\frac{1-a}{1+a}}\right) \end{aligned}

\begin{aligned} I(a) &=2 \int \frac{1}{\sqrt{1-a^{2}}} \tan ^{-1}\left(\sqrt{\frac{1-a}{1+a}}\right) d a \\ &\stackrel{a\mapsto\cos \theta}{=} -2 \int \frac{1}{\sin \theta} \tan ^{-1} \sqrt{\tan ^{2} \frac{\theta}{2}} \sin \theta d \theta \\ &=-\int \theta d \theta\\ &=-\frac{\left(\cos ^{-1} a\right)^{2}}{2}+C \end{aligned}

Taking limits $$0$$ to $$a$$, we get the final result. \begin{aligned} I(a) &=\int_{0}^{a} I^{\prime}(y) d y =\left[-\frac{\left(\cos ^{-1} y\right)^{2}}{2}\right]_{0}^{a} =\frac{\pi^{2}}{8}-\frac{\left(\cos ^{-1} a\right)^{2}}{2} \end{aligned}