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

The integral from 0 to pi of (2+cos(x))^2 is 4pi+pi/2

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

\int_{0}^{π} (2 + cos (x))^{2} d x

4 π + \frac{π}{2}

Decimal

14.13716 \dots

Solution steps

\int_{0}^{π} (2 + cos (x))^{2} d x

Expand

(2 + cos (x))^{2} : 4 + 4 cos (x) + cos^{2} (x)

= \int_{0}^{π} 4 + 4 cos (x) + 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

= \int_{0}^{π} 4 d x + \int_{0}^{π} 4 cos (x) d x + \int_{0}^{π} cos^{2} (x) d x

\int_{0}^{π} 4 d x = 4 π

\int_{0}^{π} 4 cos (x) d x = 0

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

= 4 π + 0 + \frac{π}{2}

Simplify

= 4 π + \frac{π}{2}

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What is the integral from 0 to pi of (2+cos(x))^2 ?
The integral from 0 to pi of (2+cos(x))^2 is 4pi+pi/2