1. $\int\limits_{0}^{\frac{\pi}{2}}\frac{sinx}{1+sinx}dx=\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\frac{sinx}{sin^{2}\left ( \frac{x}{2} +\frac{\pi}{4}\right )}dx$. Đặt: $\frac{x}{2}+\frac{\pi}{4}=t\Rightarrow x=2t-\frac{\pi}{2}\Rightarrow dx=2dt$.
Tích phân mới:
$\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{sin\left ( 2t-\frac{\pi}{2} \right )}{sin^{2}t}dt=-\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{cos2t}{sin^{2}t}dt=\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{2sin^{2}t-1}{sin^{2}t}dt=2\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}dt-\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{sin^{2}t}dt$
2. $\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{2+cosx}dx=\int\limits_{0}^{\frac{\pi}{2}}\frac{1}{1+2cos^{2}\frac{x}{2}}dx=\int\limits_{0}^{1}\frac{2}{3+tan^{2}\frac{x}{2}}d\left ( tan\frac{x}{2} \right )=\int\limits_{0}^{1}\frac{2}{3+t^{2}}dt$.
Đặt: $t=\sqrt{3}tanu\Rightarrow dt=\frac{\sqrt{3}}{cos^{2}u}du$.
Thu gọn đi ta được tích phân:
$\frac{2\sqrt{3}}{3}\int\limits_{0}^{\frac{\pi}{6}}du$