a.Ta có:$S(D)=\int\limits_{\pi/4}^{\pi/4}|\tan^3x|dx$ $=2\int\limits_0^{\pi/4}\tan^3xdx$ $=2\int\limits_0^{\pi/4}(\tan^3x+\tan x-\tan x)dx$ $=2\int\limits_0^{\pi/4}\tan x(\tan^2x+1)dx-2\int\limits_0^{\pi/4}\dfrac{\sin x}{\cos x}dx$ $=2\int\limits_0^{\pi/4}\tan xd(\tan x)+2\int\limits_0^{\pi/4}\dfrac{d(\cos x)}{\cos x}$ $=\tan^2x\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.+2\ln(\cos x)\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=1-\ln2$
a.Ta có:$S(D)=\int\limits_{
-\pi/4}^{\pi/4}|\tan^3x|dx$ $=2\int\limits_0^{\pi/4}\tan^3xdx$ $=2\int\limits_0^{\pi/4}(\tan^3x+\tan x-\tan x)dx$ $=2\int\limits_0^{\pi/4}\tan x(\tan^2x+1)dx-2\int\limits_0^{\pi/4}\dfrac{\sin x}{\cos x}dx$ $=2\int\limits_0^{\pi/4}\tan xd(\tan x)+2\int\limits_0^{\pi/4}\dfrac{d(\cos x)}{\cos x}$ $=\tan^2x\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.+2\ln(\cos x)\left|\begin{array}{l}\dfrac{\pi}{4}\\0\end{array}\right.=1-\ln2$