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

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

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

\int_{0}^{2 π} (cos (t))^{2} d t

π

Decimal

3.14159 \dots

Solution steps

\int_{0}^{2 π} (cos (t))^{2} d t

Simplify

= \int_{0}^{2 π} cos^{2} (t) d t

Rewrite using trig identities

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

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 t) d t

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 t + \int_{0}^{2 π} cos (2 t) d t)

\int_{0}^{2 π} 1 d t = 2 π

\int_{0}^{2 π} cos (2 t) d t = 0

= \frac{1}{2} (2 π + 0)

Simplify

\frac{1}{2} (2 π + 0) : π

= π

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