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

The integral from 0 to pi/2 of (1+cos(θ))^2 is (3pi)/4+2

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

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

\frac{3 π}{4} + 2

Solution steps

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

Expand

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

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

\int_{0}^{\frac{π}{2}} 1 d θ = \frac{π}{2}

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

\int_{0}^{\frac{π}{2}} cos^{2} (θ) d θ = \frac{π}{4}

= \frac{π}{2} + 2 + \frac{π}{4}

Simplify

\frac{π}{2} + 2 + \frac{π}{4} : \frac{3 π}{4} + 2

= \frac{3 π}{4} + 2

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What is the integral from 0 to pi/2 of (1+cos(θ))^2 ?
The integral from 0 to pi/2 of (1+cos(θ))^2 is (3pi)/4+2