Let  $$I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {{x \over {1 + \sin x}}dx} $$

also let K = $${x \over {1 + \sin x}}$$

Multiplying numerator and denominator by
(1 $$-$$ sin x) we get;

$$K = {{x\left( {1 - \sin x} \right)} \over {1 - {{\left( {\sin x} \right)}^2}}} = {{x(1 - \sin x)} \over {{{\left( {\cos x} \right)}^2}}}$$

= x(1 $$-$$ sin x) sec²x

= x sec²x $$-$$ x sin x sec²x = x sec²x $$-$$ x tan x sec x

Now, $$I = \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x{{\sec }^2}xdx - \int_{{\pi \over 4}}^{{{3\pi } \over 4}} {x\sec x\,\tan \,xdx} } $$

$$ \Rightarrow $$ I = $$\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \left[ {\log \left( {\sec x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$

- $$\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}} - \int\limits_{{\pi \over 4}}^{{{3\pi } \over 4}} {{{dx} \over {\cos x}}} $$

$$ \Rightarrow $$ I = $$\left| {x\tan x} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}$$ - $$\left| {{x \over {\cos x}}} \right|_{{\pi \over 4}}^{{{3\pi } \over 4}}$$

$$ - \left[ {\log \left( {\sec x + \tan x} \right)} \right]_{{\pi \over 4}}^{{{3\pi } \over 4}}$$

$$ \Rightarrow $$ I = $$ - {{3\pi } \over 4} - {\pi \over 4}$$$$ + {{3\pi } \over 4}\left( {\sqrt 2 } \right)$$$$ + {\pi \over 4}\left( {\sqrt 2 } \right)$$

$$ - \log \left( {\sqrt 2 + 1} \right)$$$$ + \log \left( {\sqrt 2 + 1} \right)$$

$$ \Rightarrow $$ I = -$$\pi $$ + $${\sqrt 2 \pi }$$

$$ \Rightarrow $$ I = $$\pi \left( {\sqrt 2 - 1} \right)$$