$ \left ( 1+ tanx*tan\frac{x}{2} \right )* \sin x =\sin x.(\frac{\sin x.sin x/2}{\cos x.\cos x/2}+1)= \sin x.(1+\frac{2\sin^2 x/2.\cos x/2}{(1-2\sin^2 x/2)\cos x/2})=\ sinx(1+\frac{2\sin^2 x/2}{1-2\sin^2 x/2})=\sin x.\frac{1}{1-2\sin^2 x/2}=\frac{\sin x}{\cos x}$$\Rightarrow I=\int\limits_{0}^{\pi/4}\tan x.d_{x}=-\ln (\cos x)$ (thế cận từ 0 đến $\pi/4$)$I=-\ln( \sqrt2/2)$
$ \left ( 1+ tanx*tan\frac{x}{2} \right )* \sin x =\sin x.(\frac{\sin x.sin x/2}{\cos x.\cos x/2}+1)=\ sinx(1+\frac{2\sin^2 x/2}{1-2\sin^2 x/2})=\sin x.\frac{1}{1-2\sin^2 x/2}=\frac{\sin x}{\cos x}$$\Rightarrow I=\int\limits_{0}^{\pi/4}\tan x.d_{x}=-\ln (\cos x)$ (thế cận từ 0 đến $\pi/4$)$I=-\ln( \sqrt2/2)$
$ \left ( 1+ tanx*tan\frac{x}{2} \right )* \sin x =\sin x.(\frac{\sin x.sin x/2}{\cos x.\cos x/2}+1)=
\sin x.(1+\frac{2\sin^2 x/2.\cos x/2}{(1-2\sin^2 x/2)\cos x/2})=\ sinx(1+\frac{2\sin^2 x/2}{1-2\sin^2 x/2})=\sin x.\frac{1}{1-2\sin^2 x/2}=\frac{\sin x}{\cos x}$$\Rightarrow I=\int\limits_{0}^{\pi/4}\tan x.d_{x}=-\ln (\cos x)$ (thế cận từ 0 đến $\pi/4$)$I=-\ln( \sqrt2/2)$