Đáp án:

$\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin x)dx= - \dfrac{\pi}{2}\ln2$

Giải thích các bước giải:

$\quad I =\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin x)dx$

Đặt $x = \dfrac{\pi}{2} - u$

$\Rightarrow dx = - du$

Đổi cận:

$\begin{array}{c|cc}x&0&&\dfrac{\pi}{2}\\\hline u&\dfrac{\pi}{2}&&0\end{array}$

Ta được:

$\quad I = -\displaystyle\int\limits_{\tfrac{\pi}{2}}^0\ln\left[\sin\left(\dfrac{\pi}{2} - u\right)\right]du$

$\Leftrightarrow I =\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\cos u)du$

$\Leftrightarrow I =\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\cos x)dx$

$\Leftrightarrow 2I =\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin x)dx + \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\cos x)dx$

$\Leftrightarrow 2I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin x.\cos x)dx$

$\Leftrightarrow 2I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln\left(\dfrac12\sin2x\right)dx$

$\Leftrightarrow 2I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin2x)dx - \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln2dx$

$\Leftrightarrow 2I = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin2x)dx - \dfrac{\pi}{2}\ln2\quad (*)$

Xét $I' = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin2x)dx$

Đặt $t = 2x$

$\Rightarrow dt = 2dx$

Đổi cận:

$\begin{array}{c|cc}x&0&&\dfrac{\pi}{2}\\\hline t&0&&\pi\end{array}$

Ta được:

$\quad I' = \dfrac12\displaystyle\int\limits_0^{\pi}\ln(\sin t)dt$

$\Leftrightarrow I' = \dfrac12\displaystyle\int\limits_0^{\pi}\ln(\sin x)dx$

Trên $[0;\pi]$, hàm số $y = \ln(\sin x)$ đối xứng qua $x =\dfrac{\pi}{2}$

Do đó:

$\displaystyle\int\limits_0^{\pi}\ln(\sin x)dx = 2\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin x)dx$

Ta được:

$\quad I' =\dfrac12\cdot 2\displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin x)dx$

$\Leftrightarrow I' = \displaystyle\int\limits_0^{\tfrac{\pi}{2}}\ln(\sin x)dx$

$\Leftrightarrow I' = I$

Khi đó:

$(*)\Leftrightarrow 2I = I - \dfrac{\pi}{2}\ln2$

$\Leftrightarrow I = - \dfrac{\pi}{2}\ln2$