Tìm nguyên hàm của: tan(x+ $\pi$ /3).cot(x+ $\pi$ /6)

Đáp án:

Giải thích các bước giải:

$\quad I =  \displaystyle\int\tan\left(x +\dfrac{\pi}{3}\right)\cot\left(x +\dfrac{\pi}{6}\right)dx$

$\to I = \displaystyle\int\dfrac{2\cos2x +1}{2\cos2x -1}dx$

Đặt $u = 2x$

$\to du = 2dx$

Ta được:

$\quad I = \dfrac12\displaystyle\int\dfrac{2\cos u +1}{2\cos u -1}du$

Đặt $t = \tan\dfrac u2$

$\to dt = \dfrac{1}{2\cos^2\left(\dfrac u2\right)}du$

Ta được:

$\quad I =\dfrac12\displaystyle\int\dfrac{2\cdot\dfrac{1-t^2}{t^2+1}+1}{2\cdot\dfrac{1-t^2}{t^2+1} -1}\cdot\dfrac{2}{t^2+1}dt$

$\to I = \displaystyle\int\dfrac{t^2 -3}{3t^4 +2t^2 -1}dt$

$\to I =\displaystyle\int\left(\dfrac{1}{t^2+1}+\dfrac{1}{\sqrt3t +1} + \dfrac{1}{1-\sqrt3t}\right)dt$

$\to I = \displaystyle\int\dfrac{1}{t^2+1}dt + \dfrac{1}{\sqrt3}\displaystyle\int\dfrac{d(\sqrt3t +1)}{\sqrt3t +1} - \dfrac{1}{\sqrt3}\displaystyle\int\dfrac{d(1-\sqrt3t)}{1-\sqrt3t}$

$\to I = \arctan t + \dfrac{1}{\sqrt3}\ln|\sqrt3t +1| - \dfrac{1}{\sqrt3}\ln|1-\sqrt3t| + C$

$\to I = \dfrac u2 + \dfrac{1}{\sqrt3}\ln\left|\dfrac{\sqrt3\tan\dfrac u2 +1}{1-\sqrt3\tan\dfrac u2}\right| + C$

$\to I = x +\dfrac{1}{\sqrt3}\ln\left|\dfrac{\sqrt3\tan x + 1}{1-\sqrt3\tan x}\right| + C$