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

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

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integral from 0 to x of sin(u)

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

- cos (x) + 1

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

- cos (x) + 1

= - cos (x) + 1

\int_{0}^{1} \frac{1}{81 - x ^{2}} d x

\int_{0}^{2} \frac{x}{5 + x ^{2}} d x

\int_{- 1}^{- 1} x^{3} d x

\int_{0}^{4} 2 t + 1 d t

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

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