∫ 0 π f ( x ) d x where f ( x ) = { sin ( x ) 0 ≤ x < π 2 cos ( x ) π 2 ≤ x ≤ π {\displaystyle \int _{0}^{\pi }f(x)\,dx\quad {\text{where}}\;f(x)={\begin{cases}\sin(x)&0\leq x<{\frac {\pi }{2}}\\\cos(x)&{\frac {\pi }{2}}\leq x\leq \pi \end{cases}}}
∫ 0 π f ( x ) d x = ∫ 0 π 2 f ( x ) d x + ∫ π 2 π f ( x ) d x = ∫ 0 π 2 sin ( x ) d x + ∫ π 2 π cos ( x ) d x = − cos ( x ) | 0 π 2 + sin ( x ) | π 2 π = [ − cos ( π 2 ) + cos ( 0 ) ] + [ sin ( π ) − sin ( π 2 ) ] = [ 0 + 1 ] + [ 0 − 1 ] = 0 {\displaystyle {\begin{aligned}\int _{0}^{\pi }f(x)\,dx&=\int _{0}^{\frac {\pi }{2}}f(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }f(x)\,dx=\int _{0}^{\frac {\pi }{2}}\sin(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }\cos(x)\,dx\\[2ex]&=-\cos(x){\bigg |}_{0}^{\frac {\pi }{2}}+\sin(x){\bigg |}_{\frac {\pi }{2}}^{\pi }\\[2ex]&=\left[-\cos \left({\frac {\pi }{2}}\right)+\cos(0)\right]+\left[\sin(\pi )-\sin \left({\frac {\pi }{2}}\right)\right]\\[2ex]&=[0+1]+[0-1]\\[2ex]&=0\end{aligned}}}
![{\displaystyle {\begin{aligned}\int _{0}^{\pi }f(x)\,dx&=\int _{0}^{\frac {\pi }{2}}f(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }f(x)\,dx=\int _{0}^{\frac {\pi }{2}}\sin(x)\,dx+\int _{\frac {\pi }{2}}^{\pi }\cos(x)\,dx\\[2ex]&=-\cos(x){\bigg |}_{0}^{\frac {\pi }{2}}+\sin(x){\bigg |}_{\frac {\pi }{2}}^{\pi }\\[2ex]&=\left[-\cos \left({\frac {\pi }{2}}\right)+\cos(0)\right]+\left[\sin(\pi )-\sin \left({\frac {\pi }{2}}\right)\right]\\[2ex]&=[0+1]+[0-1]\\[2ex]&=0\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/50883fffc97606b59a90fcd2e7e683cd36295bcc)