Tính tích phân $\Large \int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (\s

Tính tích phân $\Large \int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (\sin x+\cos x)}{{{\cos }^{2}}x}dx}$ , ta được kết quả

• A.

  $\Large -\dfrac{\pi }{4}+\dfrac{1}{2}\ln 2$

• B.

  $\Large \dfrac{\pi }{4}-\dfrac{3}{2}\ln 2$

• C.

  $\Large -\dfrac{\pi }{4}+\dfrac{3}{2}\ln 2$

• D.

  $\Large -\dfrac{\pi }{4}-\dfrac{3}{2}\ln 2$

Chọn C

$\Large\int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (\sin x+\cos x)}{{{\cos }^{2}}x}dx=\int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (\cos x.(1+\tan x))}{{{\cos }^{2}}x}dx}}$$\Large =\int\limits_{0}^{\dfrac{\pi}{4}}{\left( \dfrac{\ln (\cos x)}{{{\cos }^{2}}x}+\dfrac{\ln (1+\tan x)}{{{\cos }^{2}}x} \right)dx}$ $\Large=\int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (\cos x)}{{{\cos }^{2}}x}dx+\int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (1+\tan x)}{{{\cos }^{2}}x}dx=I+J}}$

+ Tính I

Đặt $\Large\left\{ \begin{align}  & u=\ln (\cos x)\Rightarrow du=-\dfrac{\sin x}{\cos x}dx \\  & dv=\dfrac{1}{{{\cos }^{2}}x}dx\Rightarrow v=\tan x \\ \end{align} \right.$

$\Large I=\int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (\cos x)}{{{\cos }^{2}}x}dx}$

$\Large =\tan x.\ln (\cos x)\left| \begin{align}  & \frac{\pi }{4} \\  & 0 \\ \end{align} \right.$ $\Large +\int\limits_{0}^{\pi /4}{{{\tan }^{2}}xdx}$

$\Large = \left. {\tan x.\ln \left( {\cos x} \right)} \right|_0^{\frac{\pi }{4}} + \int\limits_0^{\frac{\pi }{4}} {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)} dx$

$\Large =\tan x.\ln (\cos x)\left| \begin{align} & \frac{\pi }{4} \\ & 0 \\ \end{align} \right.$$\Large+(-x+\tan x)\left| \begin{align} & \frac{\pi }{4} \\ & 0 \\ \end{align} \right.$$\Large=-\dfrac{1}{2}\ln 2-\dfrac{\pi }{4}+1$

+ Tính J

$\Large J=\int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (1+\tan x)}{{{\cos }^{2}}x}dx}$. Đặt $\Large t=1+\tan x\Rightarrow dt=\dfrac{1}{{{\cos }^{2}}x}dx$

Đổi cận: $\Large x=0\Rightarrow t=1,x=\dfrac{\pi }{4}\Rightarrow t=2$

$\Large J=\int\limits_{1}^{2}{\ln tdt}$. Đặt $\Large \left\{ \begin{align}  & u=\ln t\Rightarrow du=\dfrac{1}{t}dt \\  & dv=dt\Rightarrow v=t \\ \end{align} \right.$ $\Large \Rightarrow J=\int\limits_{1}^{2}{\ln tdt=(t\ln t-t)\left| \begin{align}  & 2 \\  & 1 \\ \end{align} \right.}=2\ln 2-1$

Vậy $\Large \int\limits_{0}^{\dfrac{\pi}{4}}{\dfrac{\ln (\sin x+\cos x)}{{{\cos }^{2}}x}dx=-\dfrac{\pi }{4}+\dfrac{3}{2}\ln 2}$