$$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$$

$$ \Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$$

$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$

Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$

$$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$

$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$

$$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$

$$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$

$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$

Let $$I = \int\limits_0^\pi {xf\left( {\sin x} \right)} dx$$

$$ = \int\limits_0^\pi {\left( {\pi - x} \right)} f\left( {\sin x} \right)dx$$

$$\therefore$$ $$2I = \pi \int\limits_2^\pi {f\left( {\sin x} \right)} dx$$

$$ = \pi .2\int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$

$$\therefore$$ $$I = \pi \int\limits_0^{{\pi \over 2}} {f\left( {\sin x} \right)} dx$$

$$ \Rightarrow A = \pi $$

$$\int\limits_{ - 2}^3 {\left| {1 - {x^2}} \right|} dx = \int\limits_{ - 2}^3 {\left| {{x^2} - 1} \right|} dx$$

Now $$\left| {{x^2} - 1} \right| = \left\{ {\matrix{ {{x^2} - 1} & {if} & {x \le - 1} \cr {1 - {x^2}} & {if} & { - 1 \le x \le 1} \cr {{x^2} - 1} & {if} & {x \ge 1} \cr } } \right.$$

$$\therefore$$ Integral is

$$\int\limits_{ - 2}^{ - 1} {\left( {{x^2} - 1} \right)dx + \int\limits_{ - 1}^1 {\left( {1 - {x^2}} \right)} } dx + \int\limits_1^3 {\left( {{x^2} - 1} \right)dx} $$

$$\left[ {{{{x^3}} \over 3} - x} \right]_{ - 2}^{ - 1} + \left[ {x - {{{x^3}} \over 3}} \right]_{ - 1}^1 + \left[ {{{{x^3}} \over 3} - x} \right]_1^3$$

$$ = \left( { - {1 \over 3} + 1} \right) - \left( { - {8 \over 3} + 2} \right) + \left( {2 - {2 \over 3}} \right) + \left( {{{27} \over 3} - 3} \right) - \left( {{1 \over 3} - 1} \right)$$

$$ = {2 \over 3} + {2 \over 3} + {4 \over 3} + 6 + {2 \over 3} = {{28} \over 3}$$

$$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin \,x + \cos x} \right)}^2}} \over {\sqrt {1 + \sin 2x} }}} dx$$

We know $$\left[ {{{\left( {\sin x + \cos x} \right)}^2} = 1 + \sin 2x} \right],\,$$ So

$$I = \int\limits_0^{{\pi \over 2}} {{{{{\left( {\sin x + \cos x} \right)}^2}} \over {\left( {\sin x + \cos x} \right)}}} dx$$

$$ = \int\limits_0^{{\pi \over 2}} {\left( {\sin x + \cos x} \right)dx} $$

$$\left[ {} \right.$$ $$\sin x + \cos x > 0$$ $$\,\,if\,\,0 < x < {\pi \over 2}$$ $$\left. {} \right]$$

or $$I = \left[ { - \cos x + \sin x} \right]_0^{{\pi \over 2}} $$

= - cos $${\pi \over 2}$$ + sin $${\pi \over 2}$$ + cos 0 - sin 0

= - 0 + 1 + 1 - 0

= 2