What is the integral from 0 to 2pi of 2cos(2x+4) ?

The integral from 0 to 2pi of 2cos(2x+4) is sin(4pi+4)-sin(4)

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integral from 0 to 2pi of 2cos(2x+4)

\int_{0}^{2 π} 2 cos (2 x + 4) d x

sin (4 π + 4) - sin (4)

Decimal Notation

0

Solution steps

\int_{0}^{2 π} 2 cos (2 x + 4) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{2 π} cos (2 x + 4) d x

Apply u-substitution

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

= 2 \cdot \int_{4}^{4 π + 4} 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

= 2 \cdot \frac{1}{2} \cdot \int_{4}^{4 π + 4} cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{2} [sin (u)]_{4}^{4 π + 4}

Simplify

2 \cdot \frac{1}{2} [sin (u)]_{4}^{4 π + 4} : [sin (u)]_{4}^{4 π + 4}

= [sin (u)]_{4}^{4 π + 4}

Compute the boundaries:

sin (4 π + 4) - sin (4)

= sin (4 π + 4) - sin (4)

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What is the integral from 0 to 2pi of 2cos(2x+4) ?
The integral from 0 to 2pi of 2cos(2x+4) is sin(4pi+4)-sin(4)