What is the integral from-1 to 1 of (2cos(pi/3 x)) ?

The integral from-1 to 1 of (2cos(pi/3 x)) is (6sqrt(3))/pi

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integral from-1 to 1 of (2cos(pi/3 x))

\int_{- 1}^{1} (2 cos (\frac{π}{3} x)) d x

\frac{6 3}{π}

Decimal

3.30797 \dots

Solution steps

\int_{- 1}^{1} 2 cos (\frac{π}{3} x) d x

Take the constant out:

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

= 2 \cdot \int_{- 1}^{1} cos (\frac{π}{3} x) d x

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= 2 \cdot \int_{- 1}^{1} cos (\frac{π x}{3}) d x

Apply u-substitution

: \int_{- \frac{π}{3}}^{\frac{π}{3}} cos (u) \frac{3}{π} d u

= 2 \cdot \int_{- \frac{π}{3}}^{\frac{π}{3}} cos (u) \frac{3}{π} d u

Take the constant out:

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

= 2 \cdot \frac{3}{π} \cdot \int_{- \frac{π}{3}}^{\frac{π}{3}} cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{3}{π} [sin (u)]_{- \frac{π}{3}}^{\frac{π}{3}}

Simplify

2 \cdot \frac{3}{π} [sin (u)]_{- \frac{π}{3}}^{\frac{π}{3}} : \frac{6}{π} [sin (u)]_{- \frac{π}{3}}^{\frac{π}{3}}

= \frac{6}{π} [sin (u)]_{- \frac{π}{3}}^{\frac{π}{3}}

Compute the boundaries:

3

= \frac{6}{π} 3

Simplify

= \frac{6 3}{π}

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What is the integral from-1 to 1 of (2cos(pi/3 x)) ?
The integral from-1 to 1 of (2cos(pi/3 x)) is (6sqrt(3))/pi