$\eqalign{
& \sin 2x + 4\sin \left( {s + {\pi \over 4}} \right) + 3 = 2{\left( {\sin \left( {x + {\pi \over 4}} \right) + 1} \right)^2} = {\left( {\sin x + \cos x + \sqrt 2 } \right)^2} \cr
& \to I = {1 \over {\sqrt 2 }}\int_0^{{\pi \over 2}} {{{x\left( {\sin x - \cos x} \right)} \over {{{\left( {\sin x + \cos x + \sqrt 2 } \right)}^2}}}dx} \cr
& u = x \to du = dx \cr
& dv = {{\sin x - \cos x} \over {{{\left( {\sin x + \cos x + \sqrt 2 } \right)}^2}}}dx \to v = {1 \over {\sin x + \cos x + \sqrt 2 }} \cr
& \to I = {1 \over {\sqrt 2 }}\left( {\left. {uv} \right|_0^{{\pi \over 2}} - \int_0^{{\pi \over 2}} {{1 \over {\sin x + \cos x + \sqrt 2 }}dx} } \right) \cr
& \int {{1 \over {\sin x + \cos x + \sqrt 2 }}dx = \int {{1 \over {\sqrt 2 \sin \left( {x + {\pi \over 4}} \right) + \sqrt 2 }}dx = {1 \over {\sqrt 2 }}\int {{1 \over {\cos \left( {x - {\pi \over 4}} \right) + 1}}dx = {1 \over {\sqrt 2 }}\int {{1 \over {2{{\cos }^2}\left( {{x \over 2} - {\pi \over 8}} \right)}}dx = {1 \over {\sqrt 2 }}\tan \left( {{x \over 2} - {\pi \over 8}} \right)} } } } \cr} $
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