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

The integral from 0 to 4pi of t^2sin(t/2) is -32pi^2

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

\int_{0}^{4 π} t^{2} sin (\frac{t}{2}) d t

- 32 π^{2}

Decimal Notation

- 315.82734 \dots

Solution steps

\int_{0}^{4 π} t^{2} sin (\frac{t}{2}) d t

Apply Integration By Parts

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

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

\int - 4 t cos (\frac{t}{2}) d t = - 16 (\frac{1}{2} t sin (\frac{t}{2}) + cos (\frac{t}{2}))

= [- 2 t^{2} cos (\frac{t}{2}) - (- 16 (\frac{1}{2} t sin (\frac{t}{2}) + cos (\frac{t}{2})))]_{0}^{4 π}

Simplify

= [- 2 t^{2} cos (\frac{t}{2}) + 16 (\frac{1}{2} t sin (\frac{t}{2}) + cos (\frac{t}{2}))]_{0}^{4 π}

Compute the boundaries:

- 32 π^{2}

= - 32 π^{2}

\frac{d y}{d x} = 6 + 6 y + x + x y

n = 1 \sum \infty \frac{2 n}{n + 1}

\int 32 x^{3} - 15 x^{2} + 8 x d x

\int 2 x^{2} sin (3 x) d x

\int ln ((4 x)^{2}) d x

What is the integral from 0 to 4pi of t^2sin(t/2) ?
The integral from 0 to 4pi of t^2sin(t/2) is -32pi^2