A calculus problem by Jose Sacramento

$\large \int_{-\pi}^\pi \dfrac{\ln(1 + \sin^2 x)} {\ln(1 + \sin^2 x) + \ln(1 + \cos^2 x)} \, dx$ . As the function is even we can write the following :

$\large 2 \int_{0}^\pi \dfrac{\ln(1 + \sin^2 x)} {\ln(1 + \sin^2 x) + \ln(1 + \cos^2 x)} \, dx$

$=\large 2 \int_{0}^{\pi/2} \dfrac{\ln(1 + \sin^2 x)} {\ln(1 + \sin^2 x) + \ln(1 + \cos^2 x)} \, dx + \large 2 \int_{\pi/2}^\pi \dfrac{\ln(1 + \sin^2 x)} {\ln(1 + \sin^2 x) + \ln(1 + \cos^2 x)} \, dx$

Let $I_1=\large \int_{0}^{\pi/2} \dfrac{\ln(1 + \sin^2 x)} {\ln(1 + \sin^2 x) + \ln(1 + \cos^2 x)} \, dx$

Applying the rule : $\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$ and adding we get $I_1=\frac{\pi}{4}$

For second part $I_2=\large \int_{\pi/2}^{\pi} \dfrac{\ln(1 + \sin^2 x)} {\ln(1 + \sin^2 x) + \ln(1 + \cos^2 x)} \, dx$

Put $x=z+\pi/2$ . Then the integral becomes like $I_1$ .So, $I_2=\frac{\pi}{4}$

So, total value of integral is $(2\times \pi/4) + (2 \times \pi/4)=\pi/2 +\pi/2=\pi=\boxed{3.14}$

Similar solution as @Kushal Bose 's presented as follows.

$\begin{aligned} I & = \int_{-\pi}^\pi \frac {\ln(1+\sin^2 x)}{\ln(1+\sin^2 x) + \ln(1+\cos^2 x)} dx \quad \quad \small \color{#3D99F6} \text{As the integrand is even.} \\ & = 2 \int_0^\pi \frac {\ln(1+\sin^2 x)}{\ln(1+\sin^2 x) + \ln(1+\cos^2 x)} dx \quad \quad \small \color{#3D99F6} \text{The integrand is symmetrical about }x = \frac \pi 2 .* \\ & = 4 \int_0^\frac \pi 2 \frac {\ln(1+\sin^2 x)}{\ln(1+\sin^2 x) + \ln(1+\cos^2 x)} dx \quad \quad \small \color{#3D99F6} \text{Using } \int_a^b f(x) \ dx = \int_a^b f(a+b - x) \ dx \\ & = 2 \int_0^\frac \pi 2 \left( \frac {\ln(1+\sin^2 x)}{\ln(1+\sin^2 x) + \ln(1+\cos^2 x)} + \frac {\ln(1+\cos^2 x)}{\ln(1+\cos^2 x) + \ln(1+\sin^2 x)} \right) dx \\ & = 2 \int_0^\frac \pi 2 dx \\ & = \pi \approx \boxed{3.142} \end{aligned}$

---

$*$ Note that $\sin^2 x = \sin^2 (\pi - x)$ and $\cos^2 x = \cos^2 (\pi - x)$ , therefore, the integrand is symmetrical about $x = \frac \pi 2$ .