6.1 Areas Between Curves/23

${\begin{aligned}&\color {red}\mathbf {y=\cos(x)} &\color {royalblue}\mathbf {y=\sin(2x)} \\&x=0&x={\frac {\pi }{2}}\\\end{aligned}}$

${\begin{aligned}\cos(x)&=\sin(2x)\\x&={\frac {\pi }{2}}\\x&={\frac {\pi }{6}}\\\end{aligned}}$

$\int _{0}^{\frac {\pi }{6}}\left(\cos(x)-\sin(2x)\right)dx+\int _{\frac {\pi }{6}}^{\frac {\pi }{2}}\left(\sin(2x)-\cos(x)\right)dx={\frac {1}{4}}+{\frac {1}{4}}={\frac {2}{4}}={\frac {1}{2}}$

${\begin{aligned}\int _{0}^{\frac {\pi }{6}}\left(\cos(x)-\sin(2x)\right)dx&=\left[\sin(x)+{\frac {1}{2}}\cos(2x)\right]{\Bigg |}_{0}^{\frac {\pi }{6}}\\[2ex]&=\left[\sin({\frac {\pi }{6}})+{\frac {1}{2}}\cos({\frac {2\pi }{6}})\right]-\left[\sin(0)+{\frac {1}{2}}\cos(2(0))\right]\\[2ex]&=\left[{\frac {1}{2}}+{\frac {1}{2}}\left({\frac {1}{2}}\right)\right]-\left[0-{\frac {1}{2}}(1)\right]\\[2ex]&={\frac {1}{2}}+{\frac {1}{4}}-{\frac {1}{2}}\\&={\frac {1}{4}}\end{aligned}}$

${\begin{aligned}\int _{\frac {\pi }{6}}^{\frac {\pi }{2}}\left[\sin(2x)-\cos(x)\right]dx&=\left[-{\frac {1}{2}}\cos(2x)-\sin(x)\right]{\Bigg |}_{\frac {\pi }{6}}^{\frac {\pi }{2}}\\[2ex]&=\left[-{\frac {1}{2}}\cos({\frac {2\pi }{2}})-\sin({\frac {\pi }{2}})\right]-\left[-{\frac {1}{2}}\cos({\frac {2\pi }{6}})-\sin({\frac {\pi }{6}})\right]\\[2ex]&=\left[-{\frac {1}{2}}\left(-1\right)-1\right]-\left[-{\frac {1}{2}}\left({\frac {1}{2}}\right)-{\frac {1}{2}}\right]\\&={\frac {1}{2}}-1+{\frac {1}{4}}+{\frac {1}{2}}\\&={\frac {1}{4}}\end{aligned}}$