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

The integral from 0 to pi/4 of (3cos^2(θ))/2 is (3(pi+2))/(16)

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integral from 0 to pi/4 of (3cos^2(θ))/2

\int_{0}^{\frac{π}{4}} \frac{3 cos ^{2} ( θ )}{2} d θ

\frac{3 ( π + 2 )}{16}

Decimal

0.96404 \dots

Solution steps

\int_{0}^{\frac{π}{4}} \frac{3 cos ^{2} ( θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{3}{2} \cdot \int_{0}^{\frac{π}{4}} cos^{2} (θ) d θ

Rewrite using trig identities

= \frac{3}{2} \cdot \int_{0}^{\frac{π}{4}} \frac{1 + cos ( 2 θ )}{2} d θ

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{3}{2} \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{4}} 1 + 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

= \frac{3}{2} \cdot \frac{1}{2} (\int_{0}^{\frac{π}{4}} 1 d θ + \int_{0}^{\frac{π}{4}} cos (2 θ) d θ)

\int_{0}^{\frac{π}{4}} 1 d θ = \frac{π}{4}

\int_{0}^{\frac{π}{4}} cos (2 θ) d θ = \frac{1}{2}

= \frac{3}{2} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2})

Simplify

\frac{3}{2} \cdot \frac{1}{2} (\frac{π}{4} + \frac{1}{2}) : \frac{3 ( π + 2 )}{16}

= \frac{3 ( π + 2 )}{16}

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What is the integral from 0 to pi/4 of (3cos^2(θ))/2 ?
The integral from 0 to pi/4 of (3cos^2(θ))/2 is (3(pi+2))/(16)