integral from 2 to sin(x) of (t^2-1)

解

\int_{2}^{s i n (x)} (t^{2} - 1) d t

\frac{sin ^{3} ( x ) - 8}{3} - sin (x) + 2

解答ステップ

\int_{2}^{s i n (x)} t^{2} - 1 d t

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{2}^{s i n (x)} t^{2} d t - \int_{2}^{s i n (x)} 1 d t

\int_{2}^{s i n (x)} t^{2} d t = \frac{sin ^{3} ( x ) - 8}{3}

\int_{2}^{s i n (x)} 1 d t = sin (x) - 2

= \frac{sin ^{3} ( x ) - 8}{3} - (sin (x) - 2)

簡素化

= \frac{sin ^{3} ( x ) - 8}{3} - sin (x) + 2