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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

\int_{0}^{2 π} (cos^{2} (t)) d t

π

Decimal

3.14159 \dots

Solution steps

\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

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