Đáp án:
$D.\ 0$
Giải thích các bước giải:
$\quad I =\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\dfrac{\ln(\cos^2x)}{\cos^2x}dx$
Đặt $\begin{cases}u = \ln(\cos^2x)\\dv = \dfrac{dx}{\cos^2x}\end{cases}\Rightarrow \begin{cases}du = - 2\tan xdx\\v = \tan x\end{cases}$
Ta được:
$\quad I = \tan x\ln(\cos^2x)\Bigg|_0^{\tfrac{\pi}{4}} + 2\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\tan^2xdx$
$\Leftrightarrow I = \ln\dfrac12 + 2\displaystyle\int\limits_0^{\tfrac{\pi}{4}}\left(\dfrac{1}{\cos^2x} -1\right)dx$
$\Leftrightarrow I = - \ln2 + 2(\tan^2x - x)\Bigg|_0^{\tfrac{\pi}{4}}$
$\Leftrightarrow I = - \ln2 + 2\left(1- \dfrac{\pi}{4}\right)$
$\Leftrightarrow I = - \dfrac12\pi - \ln2 + 2$
$\Rightarrow \begin{cases}a = -\dfrac12\\b = -1\\c = 2\end{cases}$
$\Rightarrow 2a + b + c = 0$
