Tích phân

N

nguyenbahiep1

[TEX]cosx = \frac{1-tan^2(\frac{x}{2})}{1+tan^2(\frac{x}{2})} \\ 2 - cosx = \frac{1+3tan^2(\frac{x}{2})}{1+tan^2(\frac{x}{2})}[/TEX]

vậy

[TEX]\int_{0}^{\frac{\pi}{2}}.\frac{(1+tan^2(\frac{x}{2}))dx}{1+3tan^2(\frac{x}{2})} \\ u = tan (\frac{x}{2}) \Rightarrow du = \frac{1}{2}.(1+tan^2(\frac{x}{2})).dx\\ \int_{0}^{1}\frac{2.du}{1+3.u^2} \\ u = \frac{1}{\sqrt{3}}.tant \Rightarrow du = \frac{1}{\sqrt{3}}.(1+tan^2t).dt \\ \int_{0}^{\frac{\pi}{3}}.\frac{\frac{2}{\sqrt{3}}.(1+tan^2t).dt}{1+tan^2t } = \int_{0}^{\frac{\pi}{3}}.\frac{2.dt}{\sqrt{3}}= \frac{2.t}{\sqrt{3}} =\frac{2.\pi}{3.\sqrt{3}} [/TEX]
 
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N

nguyenbahiep1

câu 2

[TEX]\sqrt{e^x -1 } = u \Rightarrow u^2 +1 = e^x \\ du = \frac{e^x.dx}{2.\sqrt{e^x -1 } } \\ du = \frac{u^2+1}{2.u}.dx \Rightarrow dx = \frac{2udu}{u^2+1} [/TEX]

[TEX]\int_{0}^{1}\frac{2u^2du}{u^2+1} = \int_{0}^{1}( 2 -\frac{2}{u^2+1}) = 2u - 2 . arctan u = 2 - \frac{\pi}{2}[/TEX]
 
N

nguyenngocchaugv

nguyenbahiep1 said:
[TEX]cosx = \frac{1-tan^2(\frac{x}{2})}{1+tan^2(\frac{x}{2})} \\ 2 - cosx = \frac{1+3tan^2(\frac{x}{2})}{1+tan^2(\frac{x}{2})}[/TEX]

vậy

[TEX]\int_{0}^{\frac{\pi}{2}}.\frac{(1+tan^2(\frac{x}{2}))dx}{1+3tan^2(\frac{x}{2})} \\ u = tan (\frac{x}{2}) \Rightarrow du = \frac{1}{2}.(1+tan^2(\frac{x}{2})).dx\\ \int_{0}^{1}\frac{2.du}{1+3.u^2} \\ u = \frac{1}{\sqrt{3}}.tant \Rightarrow du = \frac{1}{\sqrt{3}}.(1+tan^2t).dt \\ \int_{0}^{\frac{\pi}{3}}.\frac{\frac{2}{\sqrt{3}}.(1+tan^2t).dt}{1+tan^2t } = \int_{0}^{\frac{\pi}{3}}.\frac{2.dt}{\sqrt{3}}= \frac{2.t}{\sqrt{3}} =\frac{2.\pi}{3.\sqrt{3}} [/TEX]
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cho hỏi, làm sao mà bạn có được hướng giải như thế. Với lại sao phải phân tích như vậy, Với cả 2 bài, mình chẳng nhìn được nó là tích phân dạng nào@-)
 
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