0 -> /4 x.( tan^{2}x + 3 ) dx 0 ->1 (3dx / 1+ x^{3} )
#1
Đã gửi 27-03-2011 - 15:36
ariesstar
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0 -> /4 x.( tan^{2}x + 3 ) dx 0 ->1 (3dx / 1+ x^{3} )
#2
Đã gửi 27-03-2011 - 17:12
câu 1: $ \int\limits_{0}^{\dfrac{\pi}{4}} x(tan^2x +3) dx =(x^2)|_{0}^{\dfrac{\pi}{4}} \int\limits_{0}^{\dfrac{\pi}{4}} x(tan^2x +1) dx =\dfrac{\pi^2}{16} + \int\limits_{0}^{\dfrac{\pi}{4}} \dfrac{x}{cos^2x} dx = \dfrac{\pi^2}{16} + (x.tanx)|_{0}^{\dfrac{\pi}{4}} - \int\limits_{0}^{\dfrac{\pi}{4}} tanx dx = \dfrac{\pi^2}{16} + \dfrac{\pi}{4} + (ln(cosx))|_{0}^{\dfrac{\pi}{4}} =\dfrac{\pi^2}{16} + \dfrac{\pi}{4} + ln\dfrac{\sqrt{2}}{2} $
#3
Đã gửi 28-03-2011 - 22:49
ariesstar
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minh van chua hieu buoc 3 lam. cau giai thich ro dc kocâu 1: $ \int\limits_{0}^{\dfrac{\pi}{4}} x(tan^2x +3) dx =(x^2)|_{0}^{\dfrac{\pi}{4}} \int\limits_{0}^{\dfrac{\pi}{4}} x(tan^2x +1) dx =\dfrac{\pi^2}{16} + \int\limits_{0}^{\dfrac{\pi}{4}} \dfrac{x}{cos^2x} dx = \dfrac{\pi^2}{16} + (x.tanx)|_{0}^{\dfrac{\pi}{4}} - \int\limits_{0}^{\dfrac{\pi}{4}} tanx dx = \dfrac{\pi^2}{16} + \dfrac{\pi}{4} + (ln(cosx))|_{0}^{\dfrac{\pi}{4}} =\dfrac{\pi^2}{16} + \dfrac{\pi}{4} + ln\dfrac{\sqrt{2}}{2} $
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