[TEX]\int_{0}^{\pi/2}\frac{x+sin^2x}{1+2sin2x}dx=\int_{0}^{\pi/2}(\frac{x}{1+2sin2x}+\frac{sin^2x}{1+2sin2x})dx=A+B[/TEX]
[TEX]A=\int_{0}^{\pi/2}\frac{x}{1+2sin2x}dx[/TEX]
\Rightarrow Đặt [TEX]x=\pi/2-t[/TEX]
[TEX]B=\int_{0}^{\pi/2}\frac{sin^2x}{1+2sin2x}dx=\int_{0}^{\pi/2}\frac{1-cos2x}{2(1+2sin2x)}dx=\int_{0}^{\pi/2}\frac{1}{2(1+2sin2x)}dx-\int_{0}^{\pi/2}\frac{cos2x}{2(1+2sin2x)}dx=C+D[/TEX]
Tính [TEX]C=\int_{0}^{\pi/2}\frac{1}{2(1+2sin2x)}dx=\int_{0}^{\pi/4}\frac{1}{2(1+2sin2x)}dx+\int_{\pi/4}^{\pi/2}\frac{1}{2(1+2sin2x)}dx=\int_{0}^{\pi/4}\frac{1}{2cos^2x(tan^2x+4tanx+1)}dx+\int_{\pi/4}^{\pi/2}\frac{1}{2sin^2x(cot^x+4cotx+1)}dx[/TEX]
cái tích phân đầu là đặt tanx, cái thứ hai là đặt cotx
Tính [TEX]D=\int_{0}^{\pi/2}\frac{cos2x}{1+2sin2x}dx[/TEX] \Rightarrow Đặt t=1+2sin2x
Đến đây là ra rồi...
Good luck to you!!!