Đáp án:
$\dfrac{3\pi}{26}- \dfrac{11}{13}\ln\dfrac23$
Giải thích các bước giải:
$\quad I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\dfrac{3\sin x -\cos x}{2\sin x + 3\cos x}dx$
$\to I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\dfrac{\cos^2x(3\tan x -1)}{(2\tan x +3)}\cdot\dfrac{1}{\cos^2x}dx$
Đặt $u = \tan x$
$\to du=\dfrac{1}{\cos^2x}dx$
Đổi cận:
$ x\quad \Big|\quad 0\qquad \dfrac{\pi}{2}$
$\overline{t\quad\,\,\Big|\quad 0\qquad +\infty}$
Ta được:
$\quad I =\displaystyle\int\limits_0^{+\infty}\dfrac{3u -1}{(u^2+1)(2u +3)}du$
$\to I = \displaystyle\int\limits_0^{+\infty}\left(\dfrac{11u +3}{13(u^2+1)} - \dfrac{22}{13(2u+3)}\right)du$
Xét $I' = \displaystyle\int\left(\dfrac{11u +3}{13(u^2+1)} - \dfrac{22}{13(2u+3)}\right)du$
$\to I' = \displaystyle\int\dfrac{88u}{26(4u^2+4)}du + \displaystyle\int\dfrac{3}{13(u^2+1)}du- \displaystyle\int\dfrac{22}{13(2u +3)}du$
$\to I' = \dfrac{11}{26}\displaystyle\int\dfrac{d(4u^2 +4)}{4u^2 + 4} + \dfrac{3}{13}\displaystyle\int\dfrac{du}{u^2+1} - \dfrac{11}{13}\displaystyle\int\dfrac{d(2u+3)}{2u+3}$
$\to I' = \dfrac{11}{26}\ln|4u^2 + 4| + \dfrac{3}{13}\arctan u - \dfrac{11}{13}\ln|2u +3| + C$
$\to I' =\dfrac{11}{26}\ln\dfrac{(4u^2+4)}{(2u+3)^2} + \dfrac{3}{13}\arctan u + C$
Khi đó:
$\quad I = \lim\limits_{t \to +\infty}\left(\dfrac{11}{26}\ln\dfrac{(4u^2+4)}{(2u+3)^2} + \dfrac{3}{13}\arctan u + C\right)\Bigg|_0^t$
$\to I = \lim\limits_{t \to +\infty}\left(\dfrac{11}{26}\ln\dfrac{(4t^2+4)}{(2t+3)^2} + \dfrac{3}{13}\arctan t\right) - \dfrac{11}{13}\ln\dfrac23$
$\to I = \dfrac{3\pi}{26}- \dfrac{11}{13}\ln\dfrac23$
