Đặt: $t=1+\cos x \Rightarrow dt=-\sin xdx$Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x\cos x}{1+\cos x}dx$$=2\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x\cos^2x}{1+\cos x}dx$$=-2\int\limits_2^1\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\left(t-2+\dfrac{1}{t}\right)dt$$=2\left(\dfrac{t^2}{2}-2t+\ln t\right)\left|\begin{array}{l}2\\1\end{array}\right.$$=2\ln2-1$
Đặt: $t=1+\cos x \Rightarrow dt=-\sin xdx$Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x\cos x}{1+\cos x}dx$$=2\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x\cos^2x}{1+\cos x}dx$$=-2\int\limits_2^1\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\left(t-2+\dfrac{1}{t}\right)dt$$=2\left(\dfrac{t^2}{2}-2t+\ln t\right)\left|\begin{array}{l}2\\1\end{array}\right.$$=2\ln2-3$
Đặt: $t=1+\cos x \Rightarrow dt=-\sin xdx$Ta có: $\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin2x\cos x}{1+\cos x}dx$$=2\int\limits_0^{\frac{\pi}{2}}\dfrac{\sin x\cos^2x}{1+\cos x}dx$$=-2\int\limits_2^1\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\dfrac{(t-1)^2}{t}dt$$=2\int\limits_1^2\left(t-2+\dfrac{1}{t}\right)dt$$=2\left(\dfrac{t^2}{2}-2t+\ln t\right)\left|\begin{array}{l}2\\1\end{array}\right.$$=2\ln2-
1$