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

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

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integral from 0 to 2pi of cos(2θ)

\int_{0}^{2 π} cos (2 θ) d θ

0

Solution steps

\int_{0}^{2 π} cos (2 θ) d θ

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

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\int_{0}^{1 - y} 2 d x

\int_{1}^{8} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

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