Em có thể làm theo hướng giải sau
[laTEX]I_1 = \int_{0}^{\frac{\pi}{4}} \frac{2xdx}{2sin^2(x+\frac{\pi}{4})} \\ \\ u = x \Rightarrow du = dx \\ \\ dv = \frac{1}{sin^2(x+\frac{\pi}{4})} \Rightarrow v = -cot(x+\frac{\pi}{4}) \\ \\ I_1 = -x.cot(x+\frac{\pi}{4}) \big|_0^{\frac{\pi}{4}}+ \int_{0}^{\frac{\pi}{4}} \frac{cos(x+\frac{\pi}{4})dx}{sin(x+\frac{\pi}{4})} \\ \\ I_1 = -x.cot(x+\frac{\pi}{4}) \big|_0^{\frac{\pi}{4}} + ln |sin(x+\frac{\pi}{4})| \big|_0^{\frac{\pi}{4}} \\ \\ \\ \\ I_2 = \int_{0}^{\frac{\pi}{4}} \frac{1+cos2x}{2(1+sin2x)}dx \\ \\ I_2 = \int_{0}^{\frac{\pi}{4}}\frac{dx}{2(1+sin2x)} + \int_{0}^{\frac{\pi}{4}}\frac{cos2xdx}{2(1+sin2x)} = I_3+I_4 \\ \\ I_4 : 1+sin2x = u \\ \\ I_3 = \int_{0}^{\frac{\pi}{4}}\frac{dx}{4sin^2(x+\frac{\pi}{4})} = -\frac{1}{4}cot(x+\frac{\pi}{4}) \big|_0^{\frac{\pi}{4}}[/laTEX]
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