$\int\limits_0^{\pi /4} {\cfrac{{{\mathop{\rm sinx}\nolimits} + cosx}}{{9 + 16\sin 2x}}} dx$

Let $I = \int\limits_0^{\pi /4} {\cfrac{{{\mathop{\rm sinx}\nolimits} + cosx}}{{9 + 16\sin 2x}}} dx$

Put $\sin x - \cos x = t$ $\Rightarrow$ $\left( {\cos x + \sin x} \right)dx = dt$

And $1 - 2\sin x\cos x = {t^2}$ $\Rightarrow$ $1 - \sin 2x = {t^2}$

When $x = \cfrac{\pi }{4},t = sin\cfrac{\pi }{4} - \cos \cfrac{\pi }{4} = \cfrac{1}{{\sqrt 2 }} - \cfrac{1}{{\sqrt 2 }} = 0$

When $x = 0,t = \sin 0 - \cos 0 = - 1$

$\therefore$ $\int\limits_0^{\pi /4} {\cfrac{{\sin x + \cos x}}{{9 + 16\sin 2x}}dx = \int\limits_{ - 1}^0 {\cfrac{{dt}}{{9 + 16\left( {1 - {t^2}} \right)}} = \int\limits_{ - 1}^0 {\cfrac{{dt}}{{25 - 16{t^2}}}} } }$

$= \cfrac{1}{{16}}\int\limits_{ - 1}^0 {\cfrac{{dt}}{{{{\left( {\cfrac{5}{4}} \right)}^2} - {t^2}}}}$

$= \cfrac{1}{{16}} \cdot \cfrac{1}{{2 \times \cfrac{5}{4}}}\left[ {\log \left| {\cfrac{{\cfrac{5}{4} + t}}{{\cfrac{5}{4} - t}}} \right|} \right]_{ - 1}^0 = \cfrac{1}{{40}}\left[ {\log 1 - \left( {\log 1 - \log 9} \right)} \right] = \cfrac{1}{{40}}\log 9$