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

The integral from 0 to 2pi of (1+sin(θ))^2 is 3pi

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integral from 0 to 2pi of (1+sin(θ))^2

\int_{0}^{2 π} (1 + sin (θ))^{2} d θ

3 π

Decimal

9.42477 \dots

Solution steps

\int_{0}^{2 π} (1 + sin (θ))^{2} d θ

Expand

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

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

\int_{0}^{2 π} 1 d θ = 2 π

\int_{0}^{2 π} 2 sin (θ) d θ = 0

\int_{0}^{2 π} sin^{2} (θ) d θ = π

= 2 π + 0 + π

Simplify

= 3 π

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What is the integral from 0 to 2pi of (1+sin(θ))^2 ?
The integral from 0 to 2pi of (1+sin(θ))^2 is 3pi