What is the integral from 0 to 4pi of t^2sin(2t) ?

The integral from 0 to 4pi of t^2sin(2t) is -8pi^2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 4pi of t^2sin(2t)

\int_{0}^{4 π} t^{2} sin (2 t) d t

- 8 π^{2}

Decimal Notation

- 78.95683 \dots

Solution steps

\int_{0}^{4 π} t^{2} sin (2 t) d t

Apply Integration By Parts

: [\frac{1}{2} (- t^{2} cos (2 t) - 2 \cdot \int - t cos (2 t) d t)]_{0}^{4 π}

= [\frac{1}{2} (- t^{2} cos (2 t) - 2 \cdot \int - t cos (2 t) d t)]_{0}^{4 π}

\int - t cos (2 t) d t = - \frac{1}{4} (2 t sin (2 t) + cos (2 t))

= [\frac{1}{2} (- t^{2} cos (2 t) - 2 (- \frac{1}{4} (2 t sin (2 t) + cos (2 t))))]_{0}^{4 π}

Simplify

[\frac{1}{2} (- t^{2} cos (2 t) - 2 (- \frac{1}{4} (2 t sin (2 t) + cos (2 t))))]_{0}^{4 π} : [\frac{1}{4} (- 2 t^{2} cos (2 t) + 2 t sin (2 t) + cos (2 t))]_{0}^{4 π}

= [\frac{1}{4} (- 2 t^{2} cos (2 t) + 2 t sin (2 t) + cos (2 t))]_{0}^{4 π}

Compute the boundaries:

\frac{- 32 π ^{2} + 1}{4} - \frac{1}{4}

= \frac{- 32 π ^{2} + 1}{4} - \frac{1}{4}

Simplify

= - 8 π^{2}

\int 2^{4 x} d x

\frac{d}{d x} (x (x - 4))

n = 0 \sum \infty \frac{n}{n + 1}

\int \frac{t}{3 t ^{2} + 4} d t

x \to 0 lim (\frac{sin ( x )}{sin ( 2 x )})

What is the integral from 0 to 4pi of t^2sin(2t) ?
The integral from 0 to 4pi of t^2sin(2t) is -8pi^2