$1/ \int\limits_{0}^{\pi}\frac{x}{sinx+1}dx$

$2/ \int\limits_{2}^{6}\frac{dx}{2x+1+\sqrt{4x+1}}$

$3/ \int\limits_{0}^{\pi/4}tanxtan(x+\frac{\pi}{3})tan(x-\frac{\pi}{3})dx$
3/ $tanx.tan(x + \frac{\pi }{3}).tan(x - \frac{\pi }{3})$
$= tanx.\frac{{tanx - \sqrt 3 }}{{1 + \sqrt 3 tanx}}.\frac{{tanx + \sqrt 3 }}{{1 - \sqrt 3 tanx}}$
$= tanx.\frac{{ta{n^2}x - 3}}{{1 - 3ta{n^2}x}}$
$= \frac{{\sin x}}{{\cos x}}\frac{{{{\sin }^2}x - 3{{\cos }^2}x}}{{{{\cos }^2}x - 3{{\sin }^2}x}}$
$ = \frac{{\sin x}}{{\cos x}}.\frac{{4{{\sin }^2}x - 3}}{{4{{\cos }^2}x - 3}}$
$= \frac{{4{{\sin }^3}x - 3\sin x}}{{4{{\cos }^3}x - 3\cos x}} = \frac{{ - \sin 3x}}{{\cos 3x}} =  - \tan 3x$
$\Rightarrow \int\limits_0^{\pi /4} t anxtan(x + \frac{\pi }{3})tan(x - \frac{\pi }{3})dx =  - \int\limits_0^{\pi /4} {\tan 3xdx} $
Do tan3x không xác định tại $x = \frac{\pi }{6} \in \left[ {0;\frac{\pi }{4}} \right]$ nên tích phân này không tồn tại.
***Chú ý: Nếu sửa cận trên thành $\frac{\pi }{9}$ hay $\frac{\pi }{12}$ gì đó thì đặt $t = \cos 3x$ là xong.

2.  $dat    t=\sqrt{4x+1} <=> t^{2}=4x+1<=> dx=\frac{tdt}{2}$
 $va  2x+1=\frac{t^{2}}{2}+\frac{1}{2}$
$tich  phan  thanh \int\limits_{3}^{5}\frac{tdt}{(t+1)^{2}}=\int\limits_{3}^{5}\frac{dt}{t+1}-\int\limits_{3}^{5}\frac{dt}{(t+1)^{2}}$
$=ln6-ln4+\frac{1}{6}-\frac{1}{4}$
1. Đặt: $t=\pi-x \Rightarrow dx=-dt$
Đổi cận: $x=0 \Rightarrow t=\pi$
               $x=\pi \Rightarrow t=0$
Suy ra:
$I=-\int\limits_\pi^0\dfrac{\pi-t}{sin(\pi-t)+1}dt$
   $=\int\limits_0^\pi\dfrac{\pi-t}{\sin t+1}dt$
   $=\pi\int\limits_0^\pi\dfrac{dt}{\sin t+1}-I$
Từ đó suy ra:
$I=\dfrac{\pi}{2}\int\limits_0^\pi\dfrac{dt}{\sin t+1}$
   $=\dfrac{\pi}{2}\left(\dfrac{2\sin\dfrac{t}{2}}{\sin\dfrac{t}{2}+\cos\dfrac{t}{2}}\right)\left|\begin{array}{l}\pi\\0\end{array}\right. =\pi$
ban ay giai cach nay hay ma cach nay minh nghj hay nhat roi –  thiensugacoi_95 11-05-13 10:05 PM
bài 1 có thể làm bằng cách khác không? –  Hello Kitty 11-05-13 09:03 PM

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