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

The integral from 0 to t of (t-x)^3sin(x) is t^3-6t+6sin(t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to t of (t-x)^3sin(x)

\int_{0}^{t} (t - x)^{3} sin (x) d x

t^{3} - 6 t + 6 sin (t)

Solution steps

\int_{0}^{t} (t - x)^{3} sin (x) d x

Expand

(t - x)^{3} sin (x) : t^{3} sin (x) - 3 t^{2} x sin (x) + 3 t x^{2} sin (x) - x^{3} sin (x)

= \int_{0}^{t} t^{3} sin (x) - 3 t^{2} x sin (x) + 3 t x^{2} sin (x) - x^{3} sin (x) d x

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}^{t} t^{3} sin (x) d x - \int_{0}^{t} 3 t^{2} x sin (x) d x + \int_{0}^{t} 3 t x^{2} sin (x) d x - \int_{0}^{t} x^{3} sin (x) d x

\int_{0}^{t} t^{3} sin (x) d x = t^{3} (- cos (t) + 1)

\int_{0}^{t} 3 t^{2} x sin (x) d x = 3 t^{2} (sin (t) - t cos (t))

\int_{0}^{t} 3 t x^{2} sin (x) d x = 3 t (- t^{2} cos (t) + 2 (t sin (t) + cos (t)) - 2)

\int_{0}^{t} x^{3} sin (x) d x = 3 (t^{2} sin (t) - 2 (sin (t) - t cos (t))) - t^{3} cos (t)

= t^{3} (- cos (t) + 1) - 3 t^{2} (sin (t) - t cos (t)) + 3 t (- t^{2} cos (t) + 2 (t sin (t) + cos (t)) - 2) - (3 (t^{2} sin (t) - 2 (sin (t) - t cos (t))) - t^{3} cos (t))

Simplify

t^{3} (- cos (t) + 1) - 3 t^{2} (sin (t) - t cos (t)) + 3 t (- t^{2} cos (t) + 2 (t sin (t) + cos (t)) - 2) - (3 (t^{2} sin (t) - 2 (sin (t) - t cos (t))) - t^{3} cos (t)) : t^{3} - 6 t + 6 sin (t)

= t^{3} - 6 t + 6 sin (t)

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

\int_{0}^{2} - 1 d x

\int_{1}^{8} 6 d x

\int_{1}^{9} \frac{5}{x} d x

\int_{0}^{16} 3 x d x

What is the integral from 0 to t of (t-x)^3sin(x) ?
The integral from 0 to t of (t-x)^3sin(x) is t^3-6t+6sin(t)