6.1 Areas Between Curves/23

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

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

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

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

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \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] &= \frac{1}{2}/left(\frac{1}{2}\right) \\ \end{align} }