Cho tích phân $I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}{\rm{d}}x}  = \dfrac{{m - \pi }}{{m + \pi }}$, giá trị của $m$ bằng :

Ta có : \(\left( {x\sin x + \cos x} \right)' = \sin x + x\cos x - \sin x = x\cos x\)

$ \Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{x^2}}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}{\rm{d}}x}  = \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{\dfrac{x}{{\cos x}}.x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}dv} $

Đặt $\left\{ \begin{array}{l}u = \dfrac{x}{{\cos x}}\\{\rm{d}}v = \dfrac{{x\cos x}}{{{{\left( {x\sin x + \cos x} \right)}^2}}}{\rm{d}}x\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{\rm{d}}u = \dfrac{{x\sin x + \cos x}}{{{{\cos }^2}x}}{\rm{d}}x\\v =  - \dfrac{1}{{x\sin x + \cos x}}\end{array} \right..$

Khi đó

$\begin{array}{l}I = \left. { - \dfrac{x}{{\cos x}}.\dfrac{1}{{x\sin x + \cos x}}} \right|_0^{\dfrac{\pi }{4}} + \int\limits_0^{\dfrac{\pi }{4}} {\dfrac{{{\rm{d}}x}}{{{{\cos }^2}x}}}  = \\ = \dfrac{{ - \dfrac{\pi }{4}}}{{\dfrac{{\sqrt 2 }}{2}}}.\dfrac{1}{{\dfrac{\pi }{4}\dfrac{{\sqrt 2 }}{2} + \dfrac{{\sqrt 2 }}{2}}} + \left. {\tan x} \right|_0^{\dfrac{\pi }{4}}\\ = \dfrac{{ - \dfrac{\pi }{4}}}{{\dfrac{1}{2}\left( {\dfrac{\pi }{4} + 1} \right)}} + 1 = \dfrac{{ - 2\pi }}{{\left( {\pi  + 4} \right)}} + 1 = \dfrac{{4 - \pi }}{{4 + \pi }} \Rightarrow m = 4\end{array}$.