[Toán 12] Tích phân khó

L

lanhnevergivesup

Last edited by a moderator:
N

nguyenbahiep1

[laTEX]\int_{0}^{\frac{\pi}{2}}( 1 + \frac{cosx}{sinx+3} - \frac{3}{sinx+3})dx = I_1+I_2 - 3I_3 [/laTEX]


[TEX]I_1[/TEX] đơn giản rồi

[laTEX]I_2: sinx +3 = u \Rightarrow du = cosx [/laTEX]

[laTEX]I_3: sinx = \frac{2tan(\frac{x}{2})}{1+tan^2(\frac{x}{2})} \\ \\ \Rightarrow \frac{1}{sinx+3} = \frac{1+tan^2(\frac{x}{2})}{3tan^2(\frac{x}{2})+2tan(\frac{x}{2}) + 3 } \\ \\ tan(\frac{x}{2}) = t [/laTEX]
 
You must log in or register to reply here.
Top Bottom