What is the integral from 0 to pi/6 of 2cos(3x) ?

The integral from 0 to pi/6 of 2cos(3x) is 2/3

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integral from 0 to pi/6 of 2cos(3x)

\int_{0}^{\frac{π}{6}} 2 cos (3 x) d x

\frac{2}{3}

Decimal Notation

0.66666 \dots

Solution steps

\int_{0}^{\frac{π}{6}} 2 cos (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_{0}^{\frac{π}{6}} cos (3 x) d x

Apply u-substitution

: \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{3} d u

= 2 \cdot \int_{0}^{\frac{π}{2}} cos (u) \frac{1}{3} d u

Take the constant out:

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

= 2 \cdot \frac{1}{3} \cdot \int_{0}^{\frac{π}{2}} cos (u) d u

Use the common integral:

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

= 2 \cdot \frac{1}{3} [sin (u)]_{0}^{\frac{π}{2}}

Simplify

2 \cdot \frac{1}{3} [sin (u)]_{0}^{\frac{π}{2}} : \frac{2}{3} [sin (u)]_{0}^{\frac{π}{2}}

= \frac{2}{3} [sin (u)]_{0}^{\frac{π}{2}}

Compute the boundaries:

1

= \frac{2}{3} \cdot 1

Simplify

= \frac{2}{3}

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What is the integral from 0 to pi/6 of 2cos(3x) ?
The integral from 0 to pi/6 of 2cos(3x) is 2/3