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[TEX]\int_{0}^{\frac{\pi }{2}}(2{cos}^{2}\frac{x}{2}+x.cosx).{e}^{sinx}.dx[/TEX]
bài này hơi khó 1 chút bạn chú ý như sau
[laTEX]I = \int_{0}^{\frac{\pi}{2}}( cosx + 1 + xcosx)e^{sinx}.dx \\ \\ I = \int_{0}^{\frac{\pi}{2}} cosx.e^{sinx}.dx + \int_{0}^{\frac{\pi}{2}}(xcosx+1)e^{sinx}.dx = I_1+I_2 \\ \\ I_1 = \int_{0}^{\frac{\pi}{2}} cosx.e^{sinx}.dx = \int_{0}^{\frac{\pi}{2}} e^{sinx}.d(sinx) = e^{sinx} \big|_0^{\frac{\pi}{2}}[/laTEX]
bây h là I_2
chú ý ta có
[laTEX]\int_{0}^{\frac{\pi}{2}}e^{sinx}.dx = x.e^{sinx} \big|_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}} xcosx.e^{sinx}.dx \\ \\ \Rightarrow \int_{0}^{\frac{\pi}{2}}e^{sinx}.dx + \int_{0}^{\frac{\pi}{2}} xcosx.e^{sinx}.dx = x.e^{sinx} \big|_{0}^{\frac{\pi}{2}} \\ \\ \Rightarrow I_2 = x.e^{sinx} \big|_{0}^{\frac{\pi}{2}} [/laTEX]
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