What is the integral from 0 to pi/9 of cos(3x) ?

The integral from 0 to pi/9 of cos(3x) is 1/(2sqrt(3))

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integral from 0 to pi/9 of cos(3x)

\int_{0}^{\frac{π}{9}} cos (3 x) d x

\frac{1}{2 3}

Decimal

0.28867 \dots

Solution steps

\int_{0}^{\frac{π}{9}} cos (3 x) d x

Apply u-substitution

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

= \int_{0}^{\frac{π}{3}} 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

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

Use the common integral:

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

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

Compute the boundaries:

\frac{3}{2}

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

Simplify

= \frac{1}{2 3}

\frac{d}{d x} (\frac{1}{2} cos (2 x))

\int (10) d x

\frac{d}{d x} (\frac{5}{( 4 - 7 x ) ^{3}})

\frac{d}{d x} (cos^{2} (x))

2 x^{2} - 8 x + 11

What is the integral from 0 to pi/9 of cos(3x) ?
The integral from 0 to pi/9 of cos(3x) is 1/(2sqrt(3))