Tính tích phân $I = \displaystyle \int_{0}^{\frac{\pi }{2}}\ln \left[ \left(1+\tan\dfrac{x}{2} \right){e}^{x} \right]dx$

Tính tích phân $I = \displaystyle \int_{0}^{\frac{\pi }{2}}\ln \left[ \left(1+\tan\dfrac{x}{2} \right){e}^{x} \right]dx$

Giải

Đặt $t=\dfrac{\pi}{2}-x$

$x=0 \Rightarrow t=\dfrac{\pi}{2} ; \,\,  x=\dfrac{\pi}{2} \Rightarrow t=0$

$I = \displaystyle \int_{0}^{\frac{\pi }{2}}\ln \left[ \left(1+\tan \left(\dfrac{\pi}{4}- \dfrac{t}{2}\right) \right){e}^{\frac{\pi}{2}-t} \right]dt$

Ta có: $\ln \left[ \left(1+\tan \left(\dfrac{\pi}{4}- \dfrac{t}{2}\right) \right){e}^{\frac{\pi}{2}-t} \right] =\ln \left[ \left(1+\dfrac{1-\tan \frac{t}{2}}{1+\tan \frac{t}{2}} \right)e^{\frac{\pi}{2}-t}\right] \\ = \ln\dfrac{2e^{\frac{\pi}{2}}}{\left(1+\tan\dfrac{t}{2}\right)e^t} = \ln 2+\dfrac{\pi}{2}-\ln \left[ \left(1+\tan\dfrac{t}{2}\right) \right]e^t$

$I= \left(\ln 2+\dfrac{\pi}{2}\right)t \bigg|_0^{\frac{\pi}{2}} - \displaystyle \int_{0}^{\frac{\pi}{2}} \ln \left[ \left(1+\tan\dfrac{x}{2}\right) \right]e^x dx=\left( \ln 2+\dfrac{\pi}{2}\right)\dfrac{\pi}{2}-I$

Vậy $I=\left(\ln 2+\dfrac{\pi}{2}\right)\dfrac{\pi}{4}$