What is the integral from 0 to t of usin(2u) ?

The integral from 0 to t of usin(2u) is 1/4 sin(2t)-(tcos(2t))/2

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integral from 0 to t of usin(2u)

\int_{0}^{t} u sin (2 u) d u

\frac{1}{4} sin (2 t) - \frac{t cos ( 2 t )}{2}

Solution steps

\int_{0}^{t} u sin (2 u) d u

Apply Integration By Parts

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

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

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

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

Simplify

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

Compute the boundaries:

\frac{1}{4} sin (2 t) - \frac{1}{2} t cos (2 t)

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

Simplify

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

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

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\int_{0}^{1} r^{2} d r

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

What is the integral from 0 to t of usin(2u) ?
The integral from 0 to t of usin(2u) is 1/4 sin(2t)-(tcos(2t))/2