## Calculus 8th Edition

$$\int_0^{\pi/2}\cos x\sin(\sin x)dx=1-\cos1$$
To evaluate the integral $$\int_0^{\pi/2}\cos x\sin(\sin x)dx$$ we will use substitution $\sin x=t$ which gives us $\cos xdx=dt$ and the integration bounds would be: for $x=0$ we have $t=0$ and for $x=\pi/2$ we have $t=1$ so we have: $$\int_0^{\pi/2}\cos x\sin(\sin x)dx=\int_0^1\sin tdt=\left.-\cos t\right|_0^1=-(\cos1-\cos0)=-(\cos1-1)=1-\cos1$$