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

The integral from 0 to t of usin(u) is sin(t)-tcos(t)

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integral from 0 to t of usin(u)

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

sin (t) - t cos (t)

Solution steps

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

Apply Integration By Parts

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

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

\int - cos (u) d u = - sin (u)

= [- u cos (u) - (- sin (u))]_{0}^{t}

Simplify

= [- u cos (u) + sin (u)]_{0}^{t}

Compute the boundaries:

sin (t) - t cos (t)

= sin (t) - t cos (t)

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

\int_{0}^{1} 4 x (x^{2} + 1)^{5} d x

\int_{4}^{7} - 9 lo g_{3} (t) d t

\int_{1}^{4} (3 x^{2} + \frac{4}{x ^{2}}) d x

\int_{1}^{4} (\frac{1}{x}) d x

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