Let's do some calculus! (19)

Case $I_1$

$\begin{aligned} I_1 & = \int_0^{\pi/2} \frac {\sin x - \cos x}{1+\sin x \cos x} dx & \small \color{#3D99F6}{\text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^{\pi/2} \left( \frac {\sin x - \cos x}{1+\sin x \cos x} + \frac {\cos x - \sin x}{1+\cos x \sin x} \right) dx \\ & = \color{#D61F06}{0 \quad \text{not the solution}} \end{aligned}$

$$ Case $I_2$

Since the integrand $\cos^6 x \ge 0$ , $\implies \color{#3D99F6}{I_2 > 0}$ . The integral can be found using beta function as follows.

$\begin{aligned} I_2 & = \int_0^{2 \pi} \cos^6 x \ dx & \small \color{#3D99F6}{\text{Since }\cos^6 x \text{ is even,}} \\ & = 4 \int_0^{\pi /2} \cos^6 x \ dx \\ & = 4 \int_0^{\pi /2} \cos^6 x \sin^0 x \ dx \\ & = 2 B \left(\frac 72, \frac 12 \right) \\ & = \frac {2 \Gamma \left(\frac 72 \right)\Gamma \left(\frac 12 \right)}{\Gamma \left(4 \right)} \\ & = \frac {2 \cdot 3 \cdot 5 \pi}{3! \cdot 2^3} \\ & = \color{#3D99F6}{\frac {5\pi}8 \ne 0 \quad \text{the solution}} \end{aligned}$

$$ Case $I_3$

Since the integrand $\sin^3 x$ is odd, $\implies \color{#D61F06}{I_3 = 0}$ as follows.

$\begin{aligned} I_3 & = \int_{-\pi/2}^{\pi/2} \sin^3 x \ dx \\ & = \int^0_{-\pi/2} \sin^3 x \ dx + \int_0^{\pi/2} \sin^3 x \ dx \\ & = \int_0^{\pi/2} \sin^3 (-x) \ dx + \int_0^{\pi/2} \sin^3 x \ dx \\ & = - \int_0^{\pi/2} \sin^3 x \ dx + \int_0^{\pi/2} \sin^3 x \ dx \\ & = \color{#D61F06}{0 \quad \text{not the solution}} \end{aligned}$

$$ Case $I_4$

$\begin{aligned} I_4 & = \int_0^1 \ln \left( \frac 1x - 1\right) dx \\ & = \int_0^1 \ln \left( \frac {1-x}x \right) dx \\ & = \int_0^1 \left(\ln(1-x)-\ln x \right) dx & \small \color{#3D99F6}{\text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx} \\ & = \frac 12 \int_0^1 \left(\ln(1-x)-\ln x + \ln(1-1+x) - \ln (1-x) \right) dx \\ & = \frac 12 \int_0^1 \left(\ln(1-x)-\ln x + \ln x - \ln (1-x) \right) dx \\ & = \color{#D61F06}{0 \quad \text{not the solution}} \end{aligned}$

$$ Answer: Therefore the answer is $\boxed{2}$ .

The reason for I2 being not 0 is clear as the function is always non negative. Others can be proved equal to 0 as in Chew-Seong's solution