integral from 0 to 1-cos(θ) of (sin(θ))r

解答

\int_{0}^{1 - c o s (θ)} (sin (θ)) r d r

\frac{sin ( θ ) ( - cos ( θ ) + 1 ) ^{2}}{2}

求解步骤

\int_{0}^{1 - c o s (θ)} sin (θ) r d r

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= sin (θ) \cdot \int_{0}^{1 - c o s (θ)} r d r

使用幂法则

= sin (θ) [\frac{r ^{2}}{2}]_{0}^{1 - c o s (θ)}

带入上下限计算后得:

\frac{( - cos ( θ ) + 1 ) ^{2}}{2}

= sin (θ) \frac{( - cos ( θ ) + 1 ) ^{2}}{2}

化简

= \frac{sin ( θ ) ( - cos ( θ ) + 1 ) ^{2}}{2}

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