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 = 1 4 + 1 4 = 2 4 = 1 2 {\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={\frac {1}{4}}+{\frac {1}{4}}={\frac {2}{4}}={\frac {1}{2}}}
∫ 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 2 ( 1 2 ) ] − [ 0 − 1 2 ( 1 ) ] = 1 2 + 1 4 − 1 2 = 1 4 {\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({\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}}}
∫ π 6 π 2 [ sin ( 2 x ) − cos ( x ) ] d x = [ − 1 2 cos ( 2 x ) − sin ( x ) ] | π 6 π 2 = [ − 1 2 cos ( 2 π 2 ) − sin ( π 2 ) ] − [ − 1 2 cos ( 2 π 6 ) − sin ( π 6 ) ] = [ − 1 2 ( − 1 ) − 1 ] − [ − 1 2 ( 1 2 ) − 1 2 ] = 1 2 − 1 + 1 4 + 1 2 = 1 4 {\displaystyle {\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}}}
![{\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({\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/97b5c8bf5fdf8b752d48d8a67086e5ec8f6b7629)
![{\displaystyle {\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}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/27b9de09cf7cbbc331aa1e674f662c733bd33c68)
