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

The integral from 0 to t of sin(x) 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(x)

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

- cos (t) + 1

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

- cos (t) + 1

= - cos (t) + 1

\int_{0}^{1} x^{4} e^{- x^{5}} d x

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

\int_{0}^{\frac{3 π}{4}} sec^{2} (x) d x

\int_{0}^{\infty} x sin (x) d x

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

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