a)\int_{0}^{pi/4}(sinx)*dx/(sinx+cosx)
câu a
[laTEX]I = \frac{1}{2}\int_{0}^{\frac{\pi}{4}}(\frac{sin x + cosx}{sin x + cosx} - \frac{cosx- sin x}{sin x + cosx})dx \\ \\ I = \frac{1}{2}\int_{0}^{\frac{\pi}{4}}( 1- \frac{cosx- sin x}{sin x + cosx})dx \\ \\ I = \frac{\pi}{8} - \frac{1}{2}\int_{0}^{\frac{\pi}{4}}\frac{d(sin x + cosx)}{sin x + cosx} \\ \\ I = \frac{\pi}{8} -\frac{1}{2} ln |sin x + cosx| \big|_0^{\frac{\pi}{4}}[/laTEX]
b)\int_{0}^{100pi}\sqrt[2]{1-cos2x}dx
....................................................
[laTEX]I = \int_{0}^{100\pi}\sqrt{2}|sinx| dx[/laTEX]
c)\int_{0}^{pi/4}(sinx+cosx)*dx/(3+sinx)
[laTEX]I = \int_{0}^{\frac{\pi}{4}} ( 1 - \frac{3}{sin x + 3} + \frac{cosx}{sin x + 3}) dx \\ \\ I = (x + ln |sin x + 3| ) \big|_0^{\frac{\pi}{4}} - 3I_1 \\ \\ I_1 = \int_{0}^{\frac{\pi}{4}} \frac{dx}{sin x+3} = \int_{0}^{\frac{\pi}{4}} \frac{(1+tan^2(\frac{x}{2}))dx}{2tan(\frac{x}{2})+3+tan^2(\frac{x}{2})} \\ \\ tan (\frac{x}{2}) = u \Rightarrow du = \frac{1}{2}(1+tan^2(\frac{x}{2})) \\ \\ I_1 = \int_{0}^{tan(\frac{\pi}{8})} \frac{2du}{u^2+2u+1+2} \\ \\ I_1 = \int_{0}^{tan(\frac{\pi}{8})}\frac{2du}{(u+1)^2 +2 } \\ \\ u = \sqrt{2}tan t [/laTEX]
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn