What is the integral from 0 to x of 1/2 sin(t) ?

The integral from 0 to x of 1/2 sin(t) is 1/2 (-cos(x)+1)

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integral from 0 to x of 1/2 sin(t)

\int_{0}^{x} \frac{1}{2} sin (t) d t

\frac{1}{2} (- cos (x) + 1)

Solution steps

\int_{0}^{x} \frac{1}{2} sin (t) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{x} sin (t) d t

Use the common integral:

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

= \frac{1}{2} [- cos (t)]_{0}^{x}

Compute the boundaries:

- cos (x) + 1

= \frac{1}{2} (- cos (x) + 1)

\int_{- 2}^{2} \frac{x ^{2}}{3} - \frac{4}{3} d x

\int_{0}^{25} - 6 x d x

\int_{0}^{\infty} x^{\frac{1}{2}} d x

\int_{0}^{1} e^{3 x + 1} d x

\int_{2}^{x} sin (x) e^{x} d x

What is the integral from 0 to x of 1/2 sin(t) ?
The integral from 0 to x of 1/2 sin(t) is 1/2 (-cos(x)+1)