What is the integral from 0 to 2pi of (1-cos(t))^2 ?

The integral from 0 to 2pi of (1-cos(t))^2 is 3pi

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integral from 0 to 2pi of (1-cos(t))^2

\int_{0}^{2 π} (1 - cos (t))^{2} d t

3 π

Decimal

9.42477 \dots

Solution steps

\int_{0}^{2 π} (1 - cos (t))^{2} d t

Expand

(1 - cos (t))^{2} : 1 - 2 cos (t) + cos^{2} (t)

= \int_{0}^{2 π} 1 - 2 cos (t) + 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

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

\int_{0}^{2 π} 1 d t = 2 π

\int_{0}^{2 π} 2 cos (t) d t = 0

\int_{0}^{2 π} cos^{2} (t) d t = π

= 2 π - 0 + π

Simplify

= 3 π

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What is the integral from 0 to 2pi of (1-cos(t))^2 ?
The integral from 0 to 2pi of (1-cos(t))^2 is 3pi