$\int\limits_{0}^{\frac{\pi }{4}} \left ( 1+ tanx*tan\frac{x}{2} \right )* \sin x  dx$
$ \left ( 1+ tanx*tan\frac{x}{2} \right )* \sin x  =\sin x.(\frac{\sin x.sin x/2}{\cos x.\cos x/2}+1)= \sin x.(1+\frac{2\sin^2 x/2.\cos x/2}{(1-2\sin^2 x/2)\cos x/2})=\ sinx(1+\frac{2\sin^2 x/2}{1-2\sin^2 x/2})=\sin x.\frac{1}{1-2\sin^2 x/2}=\frac{\sin x}{\cos x}$
$\Rightarrow I=\int\limits_{0}^{\pi/4}\tan x.d_{x}=-\ln (\cos x)$  (thế cận từ 0 đến $\pi/4$)
$I=-\ln( \sqrt2/2)$
CT vay do –  Вő˚CĦĬĈŊ♂ 21-06-13 02:41 PM
ban oi minh thay dung ma, Вő˚CĦĬĈŊ♂ dung CT sin2x =2sinx/2.cosx/2cai nay cơ bản mà –  ♂+♀→♥ 21-06-13 02:29 PM
tại sao sin x nhân sin x/2 sao lại bằng 2(sinx)^2 vậydùng công thức nào –  ngolam39 21-06-13 01:04 PM
tại lâu rồi không làm nên không nhớ –  ngolam39 21-06-13 12:57 PM
dung roi ma –  ♂+♀→♥ 21-06-13 06:40 AM
troi ạ de minh them 1 buoc –  Вő˚CĦĬĈŊ♂ 20-06-13 03:48 PM
tại sao sin x nhân sin x/2 lại bằng 2*(sinx/2)^2 vậy bạn –  ngolam39 20-06-13 02:31 PM

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