What is the integral from 0 to t of cos(2x) ?

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

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integral from 0 to t of cos(2x)

\int_{0}^{t} cos (2 x) d x

\frac{1}{2} sin (2 t)

Solution steps

\int_{0}^{t} cos (2 x) d x

Apply u-substitution

: \int_{0}^{2 t} cos (u) \frac{1}{2} d u

= \int_{0}^{2 t} cos (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 t} cos (u) d u

Use the common integral:

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

= \frac{1}{2} [sin (u)]_{0}^{2 t}

Compute the boundaries:

sin (2 t)

= \frac{1}{2} sin (2 t)

\int_{0}^{1} e^{8 x - 7} d x

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

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

\int_{- 2}^{1} - x^{2} + 9 d x

\int_{2}^{5} 2 x + 1 d x

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