[tex]I=\displaystyle \int ^{\pi}_{\tfrac{\pi}3} \dfrac{ \sin x}{1 + \cos^2 x} dx\\ \\\\\text{let,}\\\\~~~~~u=\cos x\\\\\implies \dfrac{du}{dx} =-\sin x\\ \\\implies \sin x~~ dx = -du\\\\\text{When}~~ x = \pi , ~~ u = \cos \pi = -1\\\\\text{When}~~ x = \dfrac{\pi}3 , ~~u = \cos \dfrac{\pi}3 =\dfrac 12\\ \\\\I =- \displaystyle \int ^{-1}_{\tfrac 12} \dfrac{du}{1+u^2}\\\\\\[/tex]
[tex]=\displaystyle \int ^{\tfrac 12}_{-1} \dfrac{du}{1+u^2}~~~~~~~~~~;\left[\displaystyle \int^{a}_b f(x) dx = - \displaystyle \int^{b}_a f(x) dx ,~ b < a\right]\\\\\\=\left[\tan^{-1} u \right]^{\tfrac 12}_{-1}~~~~~~~~;\left[ \ddisplaystyle \int \dfrac{dx}{ 1+ x^2} = \tan^{-1} x + C \right]\\\\\\=\tan^{-1} \left( \dfrac 12 \right) + \tan^{-1} 1\\\\\\=\tan^{-1} \left( \dfrac 12 \right) + \dfrac{\pi}4 \\\\\\=1.249[/tex]
