integral from 0 to x of t(t-sin(t))

解

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

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

解答ステップ

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

拡張

t (t - sin (t)) : t^{2} - t sin (t)

= \int_{0}^{x} t^{2} - t sin (t) 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_{0}^{x} t^{2} d t - \int_{0}^{x} t sin (t) d t

\int_{0}^{x} t^{2} d t = \frac{x ^{3}}{3}

\int_{0}^{x} t sin (t) d t = sin (x) - x cos (x)

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

簡素化

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