y = cos ( x ) y = sin ( 2 x ) x = 0 x = π 2 {\displaystyle {\begin{aligned}&\color {red}\mathbf {y=\cos(x)} &\color {royalblue}\mathbf {y=\sin(2x)} \\&x=0&x={\frac {\pi }{2}}\\\end{aligned}}}
cos ( x ) = sin ( 2 x ) x = π 2 x = π 6 {\displaystyle {\begin{aligned}\cos(x)&=\sin(2x)\\x&={\frac {\pi }{2}}\\x&={\frac {\pi }{6}}\\\end{aligned}}}
∫ 0 π 6 ( cos ( x ) − sin ( 2 x ) ) d x + ∫ π 6 π 2 ( sin ( 2 x ) − cos ( x ) ) d x {\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}
∫ 0 π 6 ( cos ( x ) − sin ( 2 x ) ) d x = [ sin ( x ) + 1 2 cos ( 2 x ) ] | 0 π 6 = [ sin ( π 6 ) + 1 2 cos ( 2 ( π 6 ) ) ] − [ sin ( 0 ) + 1 2 cos ( 2 ( 0 ) ) ] = 1 2 + 1 4 − ( 0 + 1 2 ) {\displaystyle {\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(2({\frac {\pi }{6}}))\right]-\left[\sin(0)+{\frac {1}{2}}\cos(2(0))\right]\\[2ex]&={\frac {1}{2}}+{\frac {1}{4}}-(0+{\frac {1}{2}})\end{aligned}}}
![{\displaystyle {\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(2({\frac {\pi }{6}}))\right]-\left[\sin(0)+{\frac {1}{2}}\cos(2(0))\right]\\[2ex]&={\frac {1}{2}}+{\frac {1}{4}}-(0+{\frac {1}{2}})\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/9dbe933ca693bd849ac8ade2e2e6fda45609c253)