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

The integral from 0 to t of sin(u) is -cos(t)+1

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 sin(u)

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

- cos (t) + 1

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

- cos (t) + 1

= - cos (t) + 1

\int_{0}^{2} \frac{1}{1 + 4 x} d x

\int_{0}^{π} (3 sin (x)) d x

\int_{1}^{5} \frac{3 ( ln ( x ) ) ^{2}}{x} d x

\int_{0}^{2} \frac{x ( 1 + 3 y ^{2} )}{4} d x

\int_{1}^{2} x^{2} y d y

What is the integral from 0 to t of sin(u) ?
The integral from 0 to t of sin(u) is -cos(t)+1