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

The integral from 0 to pi/6 of cos^2(3x) is pi/(12)

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

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

\frac{π}{12}

Decimal

0.26179 \dots

Solution steps

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

Rewrite using trig identities

= \int_{0}^{\frac{π}{6}} \frac{1}{2} (1 + cos (6 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} 1 + cos (6 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int_{0}^{\frac{π}{6}} 1 d x + \int_{0}^{\frac{π}{6}} cos (6 x) d x)

\int_{0}^{\frac{π}{6}} 1 d x = \frac{π}{6}

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

= \frac{1}{2} (\frac{π}{6} + 0)

Simplify

\frac{1}{2} (\frac{π}{6} + 0) : \frac{π}{12}

= \frac{π}{12}

\int cot^{3} (8 x) d x

\int \frac{3}{2} sin (x) d x

y = 40 (x^{2} - 2 x)

f (x) = 2 x^{2} - 5 x + 3, at a = 1

\frac{d}{d x} ((y^{2} + x^{2}) e^{y^{2} - x^{2}})

What is the integral from 0 to pi/6 of cos^2(3x) ?
The integral from 0 to pi/6 of cos^2(3x) is pi/(12)