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

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

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

\int_{0}^{2 π} cos^{2} (x) d x

π

Solution steps

\int_{0}^{2 π} cos^{2} (x) d x

Rewrite using trig identities

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

\int_{0}^{2 π} 1 d x = 2 π

\int_{0}^{2 π} cos (2 x) d x = 0

= \frac{1}{2} (2 π + 0)

Simplify

\frac{1}{2} (2 π + 0) : π

= π

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

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

\frac{d}{d x} (16 sin (2 x))

\frac{1}{2 s ^{2} + 4 s + 3}

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