[Toán 12] Tính tích phân

N

newstarinsky

câu 1

đặt [TEX]x=sint\Rightarrow dx=cost.dt[/TEX]

Đổi cận [TEX]x=0\Rightarrow t=0[/TEX]

[TEX]x=\frac{1}{2}\Rightarrow t=\frac{\pi}{6}[/TEX]

[tex]I=\int\limits_{0}^{\frac{\pi}{6}}\frac{sin^2t.cost}{(1-sin^2t)^2}dt[/tex]

[tex]=\int\limits_{0}^{\frac{\pi}{6}}\frac{sin^2t}{cos^3t}dt[/tex]


[tex]=\int\limits_{0}^{\frac{\pi}{6}}(\frac{1}{cos^3t}-\frac{1}{cost})dt[/tex]

[tex]=\int\limits_{0}^{\frac{\pi}{6}}[\frac{d(sint)}{(1-sin^2t)^2}-\frac{d(sint)}{1-sin^2t}][/tex]

Dạng quen rồi
 
You must log in or register to reply here.
Top Bottom