a.Ta có:S(D)=π/4∫π/4|tan3x|dx =2π/4∫0tan3xdx =2π/4∫0(tan3x+tanx−tanx)dx =2π/4∫0tanx(tan2x+1)dx−2π/4∫0sinxcosxdx =2π/4∫0tanxd(tanx)+2π/4∫0d(cosx)cosx =tan2x|π40+2ln(cosx)|π40=1−ln2
a.Ta có:$S(D)=\int\limits_{
-\pi/4}^{\pi/4}|\tan^3x|dx
=2\int\limits_0^{\pi/4}\tan^3xdx
=2\int\limits_0^{\pi/4}(\tan^3x+\tan x-\tan x)dx
=2\int\limits_0^{\pi/4}\tan x(\tan^2x+1)dx-2\int\limits_0^{\pi/4}\dfrac{\sin x}{\cos x}dx
=2\int\limits_0^{\pi/4}\tan xd(\tan x)+2\int\limits_0^{\pi/4}\dfrac{d(\cos x)}{\cos x}
=\tan^2x\left|
π40\right.+2\ln(\cos x)\left|
π40\right.=1-\ln2$