$\int_0^{2a} \frac{f(x)}{f(x) + f(2a - x)} \, dx = $

$\int_{ - \pi /2}^{\pi /2} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}dx} $ is equal to

Let $I_1 = \int\limits_0^{\frac{\pi }{2}} {{e^{ - {x^2}}}\sin (x)dx} $,$I_2 = \int\limits_0^{\frac{\pi }{2}} {{e^{ - {x^2}}}dx} $,and $I_3 = \int\limits_0^{\frac{\pi }{2}} {{e^{ - {x^2}}}(1 + x)\,dx} $. Consider the following statements:
$I: I_1 < I_2$
$II: I_2 < I_3$
$III: I_1 = I_3$
Which of the following is (are) true?

If $I$ is the greatest of the definite integrals
${I_1} = \int_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} , \,\, {I_2} = \int_0^1 {{e^{ - {x^2}}}} {\cos ^2}x\,dx$
${I_3} = \int_0^1 {{e^{ - {x^2}}}dx} ,\,\,{I_4} = \int_0^1 {{e^{ - {x^2}/2}}dx} ,$ then

$\int_{-\pi / 8}^{\pi / 8} \frac{\sin ^4(4 x)}{1+e^{4 x}} d x=$

If $\int_{-1}^4 f(x) dx = 4$ and $\int_2^4 \{3 - f(x)\} dx = 7$, then find the value of $\int_{-1}^2 f(x) dx$.