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

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

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integral from 0 to x of sin(t)

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

- cos (x) + 1

Solution steps

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

Use the common integral:

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

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

Compute the boundaries:

- cos (x) + 1

= - cos (x) + 1

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

\int_{1}^{100} \frac{1}{1 + x ^{2}} d x

4 x^{2}, 25

\frac{\partial}{\partial x} (x^{2} + 2 x y + y^{2})

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = e^{2 x}

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