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

The integral from 0 to pi of (1+cos(θ))^2 is pi+pi/2

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integral from 0 to pi of (1+cos(θ))^2

\int_{0}^{π} (1 + cos (θ))^{2} d θ

π + \frac{π}{2}

Decimal

4.71238 \dots

Solution steps

\int_{0}^{π} (1 + cos (θ))^{2} d θ

Expand

(1 + cos (θ))^{2} : 1 + 2 cos (θ) + cos^{2} (θ)

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

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} 1 d θ + \int_{0}^{π} 2 cos (θ) d θ + \int_{0}^{π} cos^{2} (θ) d θ

\int_{0}^{π} 1 d θ = π

\int_{0}^{π} 2 cos (θ) d θ = 0

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

= π + 0 + \frac{π}{2}

Simplify

= π + \frac{π}{2}

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What is the integral from 0 to pi of (1+cos(θ))^2 ?
The integral from 0 to pi of (1+cos(θ))^2 is pi+pi/2