Bạn nào giải giúp mình bài tich phân này đi
[tex]\int\limits_{0}^{\pi/2}\frac{e^x(1+sinx)}{1+cosx}dx[/tex]
[tex]I=\int\limits_{0}^{\pi/2}\frac{e^x(1+sinx)}{1+cosx}dx[/tex]
bạn đặt
[TEX]u=\frac{1+sinx}{1+cosx}--->du=\frac{1+cosx+sinx}{(1+cosx)^2}dx[/TEX]
[TEX]dv=e^xdx--->v=e^x[/TEX]
[TEX]I=\frac{1+cosx}{1+sinx}e^x+C(thaycan)- \int\limits_{0}^{\pi/2}\frac{e^x(1+cosx+sinx)}{(1+cosx)^2}dx[/TEX](
)
bay gio xet tich phan
[TEX] \int\limits_{0}^{\pi/2}\frac{e^x(1+cosx+sinx)}{(1+cosx)^2}dx[/TEX]
tach thanh hai tich phan
[TEX]\int\limits_{0}^{\pi/2}\frac{e^x)}{(1+cosx)}dx+ \int\limits_{0}^{\pi/2}\frac{e^xsinx}{(1+cosx)^2}dx=I1+I2 (*)[/tex]
xets tich phan I1
[tex]\int\limits_{0}^{\pi/2}\frac{e^x}{(1+cosx)}dx[/tex]
bạn đặt
[TEX]u=\frac{1}{1+cosx}--->du=\frac{-sinx}{(1+cosx)^2}dx[/TEX]
[TEX]dv=e^xdx--->v=e^x[/TEX]
[TEX]I1=\frac{e^x}{1+cosx}+C(thay can)+\int\limits_{0}^{\pi/2}\frac{e^xsinx)}{(1+cosx)^2}dx[/TEX]
thay vao (*)
I1+I2=[TEX]\frac{e^x}{1+cosx}+C(thay can)+\int\limits_{0}^{\pi/2}\frac{2e^xsinx)}{(1+cosx)^2}dx[/TEX](**)
tiep tuc xet tich phan
[TEX]I2=\int\limits_{0}^{\pi/2}\frac{2e^xsinx)}{(1+cosx)^2}dx[/TEX]
ban dat
[TEX]u=e^x--->du=e^xdx[/TEX]
[TEX]dv=\frac{2sinx}{(1+cosx)^2}dx--->v=\frac{2}{(1+cosx)}[/TEX]
[TEX]\frac{e^x}{(1+cosx)}+C(thay can)-(\int\limits_{}^{}\frac{2e^x}{(1+cosx)}dx=I1)[/TEX]
[TEX]I2=\frac{\frac{e^x}{(1+cosx)}+C(thay can)-\frac{e^x}{1+cosx}+C(thay can)}{2}=0[/TEX]
[TEX]I1+I2=\frac{e^x}{1+cosx}+C[/tex]
thay vao
thay can la ok
no hoi rac roi
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