6.1 Areas Between Curves/23

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

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

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 \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 }

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