What is the integral from 0 to pi/2 of (8cos^3(θ))/3 ?

The integral from 0 to pi/2 of (8cos^3(θ))/3 is 16/9

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integral from 0 to pi/2 of (8cos^3(θ))/3

\int_{0}^{\frac{π}{2}} \frac{8 cos ^{3} ( θ )}{3} d θ

\frac{16}{9}

Decimal

1.77777 \dots

Solution steps

\int_{0}^{\frac{π}{2}} \frac{8 cos ^{3} ( θ )}{3} d θ

Take the constant out:

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

= \frac{8}{3} \cdot \int_{0}^{\frac{π}{2}} cos^{3} (θ) d θ

Rewrite using trig identities

= \frac{8}{3} \cdot \int_{0}^{\frac{π}{2}} (1 - sin^{2} (θ)) cos (θ) d θ

Apply u-substitution

: \int_{0}^{1} 1 - u^{2} d u

= \frac{8}{3} \cdot \int_{0}^{1} 1 - u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{8}{3} (\int_{0}^{1} 1 d u - \int_{0}^{1} u^{2} d u)

\int_{0}^{1} 1 d u = 1

\int_{0}^{1} u^{2} d u = \frac{1}{3}

= \frac{8}{3} (1 - \frac{1}{3})

Simplify

\frac{8}{3} (1 - \frac{1}{3}) : \frac{16}{9}

= \frac{16}{9}

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What is the integral from 0 to pi/2 of (8cos^3(θ))/3 ?
The integral from 0 to pi/2 of (8cos^3(θ))/3 is 16/9