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

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

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

\int_{0}^{2 π} cos^{2} (6 x) d x

π

Decimal Notation

3.14159 \dots

Solution steps

\int_{0}^{2 π} cos^{2} (6 x) d x

Rewrite using trig identities

= \int_{0}^{2 π} \frac{1}{2} (1 + cos (12 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}^{2 π} 1 + cos (12 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}^{2 π} 1 d x + \int_{0}^{2 π} cos (12 x) d x)

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (12 x) d x = 0

= \frac{1}{2} (2 π + 0)

Simplify

\frac{1}{2} (2 π + 0) : π

= π

\frac{\partial}{\partial x} (2 x^{3} + 3 y^{2} - 6 x y + 2)

\frac{d}{d x} ((2 x^{4} - x + 1) (- x^{5} + 4))

\int 5 x^{9} d x

4 t^{\frac{1}{2}}

4 - x^{2}, 2 - x

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