Xet tich phan $I=\int\limits_{0}^{\frac{\pi }{2}}\frac{sinx}{\sqrt{5}sinx+cosx}d$
$J=\int\limits_{0}^{\frac{\pi }{2}}\frac{cosx}{\sqrt{5}sinx+cosx}dx$
Co $\sqrt{5}I+J=\int\limits_{0}^{\frac{\pi }{2}}dx=\frac{\pi}{2}$
Co $I-\sqrt{5}J=\int\limits_{0}^{\frac{\pi}{2}}\frac{sinx-\sqrt{5}cosx}{\sqrt{5}sinx+cosx}dx=-\int\limits_{0}^{\frac{\pi}{2}}\frac{d(\sqrt{5}sinx+cosx)}{\sqrt{5}sinx+cosx}=-ln\left| {\sqrt{5}sinx+cosx} \right|\begin{cases}\frac{\pi}{2} \\ 0 \end{cases}=-ln\sqrt{5}$
Sau do la giai he pt 2 an $I va J$ thoi
Den day chac ban cung lam tiep dk roi