What is the integral from 0 to \sqrt[3]{pi^2 of}sqrt(θ)cos^2(θ^{3/2}) ?

The integral from 0 to \sqrt[3]{pi^2 of}sqrt(θ)cos^2(θ^{3/2}) is pi/3

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integral from 0 to \sqrt[3]{pi^2 of}sqrt(θ)cos^2(θ^{3/2})

\frac{π}{3}

Solution steps

Rewrite using trig identities

Take the constant out:

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

Apply u-substitution

: \int_{0}^{2 π} \frac{1 + cos ( u )}{3} d u

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

Take the constant out:

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

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

\int_{0}^{2 π} 1 d u = 2 π

\int_{0}^{2 π} cos (u) d u = 0

= \frac{1}{2} \cdot \frac{1}{3} (2 π + 0)

Simplify

\frac{1}{2} \cdot \frac{1}{3} (2 π + 0) : \frac{π}{3}

= \frac{π}{3}

\frac{\partial}{\partial x} (\frac{2}{x} - x)

\frac{d}{d x} (- 8 e^{- 2 x})

x \to 4 lim (2 \cdot e^{\frac{2}{x - 4}})

x \to 0 lim ((- 3) sin (7 x))

y^{^{''}} - 1 = 0

What is the integral from 0 to \sqrt[3]{pi^2 of}sqrt(θ)cos^2(θ^{3/2}) ?
The integral from 0 to \sqrt[3]{pi^2 of}sqrt(θ)cos^2(θ^{3/2}) is pi/3