giup minh tim nguyen ham nay voi

A

alizeeduong

$2/.$ Đặt $x=\cos 2t$ với $t\in \left[0;\dfrac{\pi}{2}\right)$, ta có $dx=-2\sin 2t dt,t=\dfrac{1}{2}\arccos x$
$$\int \sqrt{\dfrac{1-x}{1+x}}dx=\int \sqrt{\dfrac{1-\cos 2t}{1+\cos 2t}}(-2\sin 2t)dt=-4\int \sin^2tdt \\ = -2\int (1-\cos 2t)dt=-2t+sìnt+\text{C}=-\arccos x+\sqrt{1-x^2}+\text{C}$$

...:)....​


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