What is the integral from 0 to x of t(t-sin(t)) ?

The integral from 0 to x of t(t-sin(t)) is (x^3)/3-sin(x)+xcos(x)

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

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

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

Solution steps

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

Expand

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

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

Apply the Sum Rule:

\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))

Simplify

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

\int_{0}^{4} x (x^{2} - 1) d x

\int_{0}^{4} \frac{1}{6 x + 1} d x

\int_{3}^{- 3} (6 x + 5)^{2} d x

\int_{0}^{1} 5^{2 t - 2} d t

\int_{0}^{b} e^{- s t} d t

What is the integral from 0 to x of t(t-sin(t)) ?
The integral from 0 to x of t(t-sin(t)) is (x^3)/3-sin(x)+xcos(x)