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

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

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integral from 0 to y of sin(x)

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

- cos (y) + 1

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

- cos (y) + 1

= - cos (y) + 1

\int_{0}^{1} cos (π) t d t

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

\int_{- 1}^{2} x + ∣ x ∣ d x

\int_{- π}^{0} (x + π) cos (n x) d x

\int_{0}^{\frac{x}{2}} \frac{sin ( x )}{x} d x

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