Question

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

Explanation:

Step1: Substitute \(u = \sin(x)\)

Differentiate \(u=\sin(x)\), \(du=\cos(x)dx\). 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 \(8\int_{0}^{1}\sin(u)du=8[-\cos(u)]_{0}^{1}\).

Step3: Evaluate the definite integral

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Answer:

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