integral from 0 to x of sqrt(1-sin(x))

解答

\int_{0}^{x} 1 - sin (x) d x

2 sin (x) + 1 - 2

求解步骤

\int_{0}^{x} 1 - sin (x) d x

使用换元积分法

= \int_{0}^{s i n (x)} \frac{1}{u + 1} d u

使用换元积分法

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

化简

\frac{1}{v} : \frac{1}{v}

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

化简

= \int_{1}^{s i n (x) + 1} \frac{1}{v} d v

使用幂法则

= [2 v^{\frac{1}{2}}]_{1}^{s i n (x) + 1}

使用根式运算法则:

= [2 v]_{1}^{s i n (x) + 1}

带入上下限计算后得:

2 sin (x) + 1 - 2

= 2 sin (x) + 1 - 2