Giúp em bài toàn này.

N

nguyenbahiep1

[laTEX]I = \int tan^3x.dx = \int (tan^3x + tanx)dx - \int tanx.dx = I_1-I_2 \\ \\ I_1 = \int tanx(tan^2x+1)dx \\ \\ tan x = u \Rightarrow du = (1+tan^2x)dx \\ \\ I_1 = \int u.du = \frac{u^2}{2} = \frac{tan^2x}{2} \\ \\ I_2 = \int \frac{sinxdx}{cosx} \\ \\ cosx = u \Rightarrow -du = sinxdx \\ \\ I_2 = \int \frac{-du}{u} = - ln|u| = -ln|cosx| \\ \\ \Rightarrow I = \frac{tan^2x}{2}+ln|cosx| +C[/laTEX]
 
S

sang115

Giải
đặt: [TEX]I = \int_{}^{}e^xcosxdx[/TEX]
[TEX]u = e^x \Rightarrow du = e^xdx[/TEX]
[TEX]dv = coxdx \Rightarrow v = sinx[/TEX]
[TEX]I = e^x.sinx - \int_{}^{}sinx.e^xdx=e^x.sinx - I_1[/TEX]
ta có: [TEX]I_1 = \int_{}^{}sinx.e^xdx[/TEX]
đặt
[TEX]u = e^x \Rightarrow du = e^xdx[/TEX]
[TEX]dv = sinxdx \Rightarrow v = -cosx[/TEX]
[TEX]I_1 = -e^x.cosx + \int_{}^{}cosx.e^xdx = -e^xcosx + I[/TEX]
[TEX]\Rightarrow I = e^x.sinx - I_1 = e^xsinx + e^xcosx - I \Leftrightarrow I = \frac{e^x}{2}(sinx + cosx) + c[/TEX]
 
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