evaluate the definite integral.\n int _ { 0 } ^ { pi / 2 } 8 cos ( x ) sin ( sin ( x ) ) d x

Explanation:

Step1: Use substitution

Let (u = \sin(x)), then (du=\cos(x)dx). When (x = 0), (u=\sin(0) = 0); when (x=\frac{\pi}{2}), (u=\sin(\frac{\pi}{2})=1). The integral (\int_{0}^{\frac{\pi}{2}}8\cos(x)\sin(\sin(x))dx) becomes (8\int_{0}^{1}\sin(u)du).

Step2: Integrate (\sin(u))

The antiderivative of (\sin(u)) is (-\cos(u)). So (8\int_{0}^{1}\sin(u)du=8\left[-\cos(u)\right]_{0}^{1}).

Step3: Evaluate the definite integral

(8\left(-\cos(1)+\cos(0)\right)=8(1 - \cos(1))).

Answer:

(8(1-\cos(1)))