$1/ \int\limits_{\pi/3}^{\pi/2}\frac{3x.cosx+2}{1+cot^2x}dx$

$2/ \int\limits_{1}^{e}\frac{lnx-2}{xlnx+x}dx$

$3/ \int\limits_{0}^{\pi/4}\frac{3cos2x-sin2x+2}{3sin2x+cos2x+3}dx$
giai bai 3 luon di Pooh. sao minh tinh khong duoc vay? –  Hello Kitty 11-05-13 03:29 PM
không bạn đưa cả tử và mẫu về cung 1 sau đó chia cho cosx^2 –  Pooh 10-05-13 10:39 PM
la dat t=tanx ha? ban nay noi ro hon duoc khong? –  Hello Kitty 10-05-13 10:37 PM
bai 3 ban nay da co y tuong chua? –  Hello Kitty 10-05-13 10:37 PM
Câu 3 bậc tử và mẫu đều là bậc hai đưa về tanx nhé –  Pooh 10-05-13 10:35 PM
$1. I=\int\limits_{\frac{\pi}3}^{\frac{\pi}2}sin^2x(2xcosx+2)dx$
nhân ra cái 1 từng phần cái hai hạ bậc
$2.I=\int\limits_{1}^{e}\frac{lnx+2-4}{xlnx+x}dx$
ta có $(xlnx+x)'=lnx+2$
cái còn lại cho $\frac1x$ vào vi phân $dlnx$
Bạn tách
$3cos2x-sin2x+2=\alpha(3sin2x+cos2x+3)+\beta(6cos2x-2sin2x)+\gamma$
$=>\alpha=0            \beta=0,5          \gamma=2$
$3.I=\frac12\int\limits_{0}^{\frac{\pi}4}\frac{d(3sin2x+cos2x+3)}{3sin2x+cos2x+3}+2\int\limits_{0}^{\frac{\pi}4}\frac1{2sin^2x+6sinxcosx+4cos^2x}dx$
đối với tích phân thứ nhất là dễ xử lí
với tích phân thứ hai bạn chia cả tử và mẫu cho $cos^2x$ dưa về $tanx$

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