Tính tích phân $I=\int_0^{\frac{\pi}{2}} \frac{1+\sin x}{1+\cos x}e^x dx $

Tính tích phân $I=\displaystyle \int_0^{\frac{\pi}{2}} \dfrac{1+\sin x}{1+\cos x}e^x dx $


Ta có: $\left( \dfrac{1+\sin x}{1+\cos x} \right ) e^x =  \left( \dfrac{1+\sin x}{ 2\cos^2 \frac{x}{2}} \right) e^x = \dfrac{e^x }{2\cos^2 \frac{x}{2}} + \dfrac{e^x \sin \frac{x}{2}\cos \frac{x}{2}}{\cos^2 \frac{x}{2}} =\dfrac{e^x }{2\cos^2 \frac{x}{2}} + e^x .\tan \frac{x}{2}$

$I=\displaystyle \int_0^{\frac{\pi}{2}} \dfrac{e^x }{2\cos^2 \frac{x}{2}}dx + \displaystyle \int_0^{\frac{\pi}{2}} e^x .\tan \frac{x}{2} dx=I_1+I_2$

Tính $I_1=\displaystyle \int_0^{\frac{\pi}{2}} \dfrac{e^x }{2\cos^2 \frac{x}{2}}dx$

Đặt
$\begin{matrix} u=e^x & \Rightarrow du=e^xdx \\ dv= \dfrac{1}{2\cos^2 \frac{x}{2}}dx &\Rightarrow v=\tan \frac{x}{2} \end{matrix}$

$I_1= e^x \tan \frac{x}{2} \bigg|_{0}^{\frac{\pi}{2}} -I_2=e^{\frac{\pi}{2}}-I_2$

Vậy $I=e^{\frac{\pi}{2}}$