6.1 Areas Between Curves/15

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

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

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

$\int _{-{\frac {\pi }{3}}}^{\frac {\pi }{3}}\left[(\tan(x))-(2\sin(x))\right]dx=\int _{-{\frac {\pi }{3}}}^{0}\left[(\tan(x))-(2\sin(x))\right]dx+\int _{0}^{\frac {\pi }{3}}\left[(2\sin(x))-(\tan(x))\right]dx=2-\ln(2)-1-1-\ln(2)+2=-2\ln(2)-2+4=-2\ln(2)+2$

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

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

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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} u &= \cos(x) \\[2ex] du &= -\sin(x) \\[2ex] -du &= \sin(x)dx \end{align} }

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://en.wikipedia.org/api/rest_v1/":): {\displaystyle \begin{align} -\int (\frac{1}{u})dx &= -\ln|u|+C &=\ln|\cos(x)^{-1}|+C &=\ln|\frac{1}{cos(x)}|+C &=\ln|\sec(x)|+C \end{align} }