$a)$ $\begin{gathered} I:=\overset{\pi }{\underset{0}{\int }}\frac{x\sin x}{1+\cos ^{2}x}dx=\overset {\pi }{\underset{0}{\int }}\frac{(\pi -x)\sin (\pi -x)}{1+\cos ^{2}(\pi -x)} dx=\overset{\pi }{\underset{0}{\int }}\frac{(\pi -x)\sin x}{1+\cos ^{2}x} dx\\ =\pi \overset{\pi }{\underset{0}{\int }}\frac{\sin x}{1+\cos ^{2}x}dx-I \end{gathered}$

$\begin{gathered} I =\frac{\pi }{2}\overset{\pi }{\underset{0}{\int }}\frac{\sin x}{1+\cos ^{2}x}dx=-\frac{\pi }{2}\overset{\pi }{\underset{0}{\int }}\frac{(\cos x)^{\prime }}{1+\cos ^{2}x}dx=-\frac{\pi }{2}\overset{\varphi (\pi )}{ \underset{\varphi (0)}{\int }}\frac{1}{1+t^{2}}dt\\ =-\frac{\pi }{2}\overset{-1} {\underset{1}{\int }}\frac{1}{1+t^{2}}dt=\frac{\pi }{2}\overset{1}{\underset{-1}{\int }}\frac{1}{1+t^{2}}dt=\frac{ \pi ^{2}}{4}. \end{gathered}$

où $\varphi :[0,\pi ]\rightarrow \lbrack -1,1]$ , $\varphi (x)=\cos x$ est une fonction de classe $C^{1}$ .

$b)$ $\begin{gathered} J :=\overset{\pi /4}{\underset{0}{\int }}\log (1+\tan x)dx=\overset{\pi /4}{ \underset{0}{\int }}\log \left(1+\tan (\frac{\pi }{4}-x)\right)dx\\ =\overset{\pi /4}{ \underset{0}{\int }}\log \left(1+\frac{1-\tan x}{1+\tan x}\right)dx =\overset{\pi /4}{\underset {0}{\int }}\log \left(\frac{2}{1+\tan x}\right)dx =\frac{\pi }{4}\log 2-J \end{gathered}$

d’où la valeur de l’intégrale est $J=\frac{\pi }{8}\log 2.$

exo7 5924