tích phân

T

truongduong9083

Câu 2

[TEX]I = \int_{}^{}\frac{1+sinx}{1+cosx}e^xdx [/TEX]
[TEX] = \int_{}^{}\frac{(sin\frac{x}{2}+cos\frac{x}{2})^2}{2cos^2 \frac{x}{2}}e^xdx[/TEX]
[TEX] =\frac{1}{2}\int_{}^{}\frac{e^xdx}{cos^2 \frac{x}{2}}+\int_{}^{}tan\frac{x}{2}e^xdx = I_1+I_2[/TEX]
Tính
[TEX]I_1 = \frac{1}{2}\int_{}^{}\frac{e^xdx}{cos^2 \frac{x}{2}}[/TEX]
Đặt
[tex]\left\{ \begin{array}{l} u = e^x \\ dv =\frac{dx}{cos^2 \frac{x}{2}}\end{array} \right.[/tex]
[tex] \Rightarrow \left\{ \begin{array}{l} du = e^xdx \\ v = 2tan\frac{x}{2} \end{array} \right.[/tex]
Nên
[TEX]I_1 =e^x.tan\frac{x}{2} - \int_{}^{}tan\frac{x}{2}e^xdx = e^x.tan\frac{x}{2} - I_2 [/TEX]
Nên [TEX]I = e^x.tan\frac{x}{2}[/TEX]. Đến đây bạn tự làm tiếp nhé
 
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K

kina31193

[TEX]I = \int_{}^{}\frac{1+sinx}{1+cosx}e^xdx [/TEX]
[TEX] = \int_{}^{}\frac{(sin\frac{x}{2}+cos\frac{x}{2})^2}{2cos^2 \frac{x}{2}}e^xdx[/TEX]
bạn à, bạn có thể giải thích giúp mình vế sau làm sao ra tg x/2 không, mình chậm hiểu, mong bạn chỉ rõ:)
[TEX] =\frac{1}{2}\int_{}^{}\frac{e^xdx}{cos^2 \frac{x}{2}}+\int_{}^{}tan\frac{x}{2}e^xdx = I_1+I_2[/TEX]
 
T

truongduong9083

[TEX]\frac{(sin\frac{x}{2}+cos \frac{x}{2})^2}{cos^2 \frac{x}{2}} = \frac{1}{2cos^2\frac{x}{2}}+\frac{2sin\frac{x}{2}.cos\frac{x}{2}}{2cos^2\frac{x}{2}}[/TEX]
[TEX]= \frac{1}{2cos^2\frac{x}{2}} + tan\frac{x}{2}[/TEX]
 
2

211666

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