6.1 Areas Between Curves/15

${\begin{aligned}&\color {red}\mathbf {y=\tan(x)} &\color {royalblue}\mathbf {y=2\sin(x)} \\&x=-{\frac {\pi }{3}}&x={\frac {\pi }{3}}\\\end{aligned}}$

$\int _{-{\frac {\pi }{3}}}^{\frac {\pi }{3}}\left[(\tan(x))-(2\sin(x))\right]dx$

${\begin{aligned}\tan(x)&=2\sin(x)\\\tan(x)-2\sin(x)&=0\\x&=0\\\end{aligned}}$

$\int _{-{\frac {\pi }{3}}}^{\frac {\pi }{3}}\left[(\tan(x))-(2\sin(x))\right]dx=\int _{-{\frac {\pi }{3}}}^{0}\left[(\tan(x))-(2\sin(x))\right]dx+\int _{0}^{\frac {\pi }{3}}\left[(2\sin(x))-(\tan(x))\right]dx={\frac {14}{3}}+{\frac {64}{3}}+{\frac {14}{3}}={\frac {92}{3}}$

${\begin{aligned}\int _{-{\frac {\pi }{3}}}^{0}\left[(\tan(x))-(2\sin(x))\right]dx\\[2ex]&=\left[\ln |sec(x)|+2\cos(x)\right]{\Bigg |}_{-{\frac {\pi }{3}}}^{0}\\[2ex]&=\left[\ln |sec(0)|+2\cos(0)\right]-\left[\ln |sec(-{\frac {\pi }{3}})+2\cos(-{\frac {\pi }{3}})|\right]\\[2ex]&=\left[0+2\right]-\left[\ln(2)-2({\frac {1}{2}})\right]=-2\ln(2)-1\\[2ex]&=-2\ln(2)-1\end{aligned}}$

${\begin{aligned}\int _{-2}^{2}\left((8-x^{2})-(x^{2})\right)dx&=\int _{-2}^{2}\left(8-2x^{2}\right)dx\\[2ex]&=\left[8x-{\frac {2x^{3}}{3}}\right]{\Bigg |}_{-2}^{2}\\[2ex]&=\left[8(2)-{\frac {2(2)^{3}}{3}}\right]-\left[8(-2)-{\frac {2(-2)^{3}}{3}}\right]\\[2ex]&=\left[16-{\frac {16}{3}}\right]-\left[-16+{\frac {16}{3}}\right]=32-{\frac {32}{3}}\\[2ex]&={\frac {64}{3}}\end{aligned}}$

${\text{Note: }}\int \tan(x)dx=\int {\frac {\sin(x)}{\cos(x)}}dx=\ln |\sec(x)|+C$

${\begin{aligned}u&=\cos(x)\\[2ex]du&=-\sin(x)\\[2ex]-du&=\sin(x)dx\end{aligned}}$

${\begin{aligned}-\int ({\frac {1}{u}})dx&=-\ln |u|+C&=\ln |\cos(x)^{-1}|+C&=\ln |{\frac {1}{cos(x)}}|+C&=\ln |\sec(x)|+C\end{aligned}}$