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

The integral from 0 to 2pi of cos(2x) is 0

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integral from 0 to 2pi of cos(2x)

\int_{0}^{2 π} cos (2 x) d x

0

Solution steps

\int_{0}^{2 π} cos (2 x) d x

Apply u-substitution

: \int_{0}^{4 π} cos (u) \frac{1}{2} d u

= \int_{0}^{4 π} 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}^{4 π} cos (u) d u

Use the common integral:

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

= \frac{1}{2} [sin (u)]_{0}^{4 π}

Compute the boundaries:

0

= \frac{1}{2} \cdot 0

Simplify

= 0

f (x) = 32 - x

y^{^{'}} - 2 \cdot y = e^{(2 \cdot t)}

\frac{d}{d x} (\frac{3 x + 2}{4 x})

\frac{d}{d x} (- sin (7 x^{2}) \cdot 14 x)

\frac{d x}{d t} + 2 t x = t, x (0) = \frac{3}{2}

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