What is the integral of θ\sqrt[4]{1-θ^2} ?

The integral of θ\sqrt[4]{1-θ^2} is -2/5 (-θ^2+1)^{5/4}+C

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integral of θ\sqrt[4]{1-θ^2}

\int θ 4 1 - θ^{2} d θ

- \frac{2}{5} (- θ^{2} + 1)^{\frac{5}{4}} + C

Solution steps

\int θ 4 1 - θ^{2} d θ

Apply u-substitution

: \int - \frac{4 u}{2} d u

= \int - \frac{4 u}{2} d u

Take the constant out:

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

= - \frac{1}{2} \cdot \int 4 u d u

Apply the Power Rule

: \frac{4}{5} u^{\frac{5}{4}}

= - \frac{1}{2} \cdot \frac{4}{5} u^{\frac{5}{4}}

Substitute back

u = 1 - θ^{2}

= - \frac{1}{2} \cdot \frac{4}{5} (1 - θ^{2})^{\frac{5}{4}}

Simplify

- \frac{1}{2} \cdot \frac{4}{5} (1 - θ^{2})^{\frac{5}{4}} : - \frac{2}{5} (- θ^{2} + 1)^{\frac{5}{4}}

= - \frac{2}{5} (- θ^{2} + 1)^{\frac{5}{4}}

Add a constant to the solution

= - \frac{2}{5} (- θ^{2} + 1)^{\frac{5}{4}} + C

\frac{1}{s ^{2} + 9}

\frac{d}{d x} (4 x - \frac{3}{x})

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

\frac{\partial}{\partial x} (sin (\frac{x}{y}))

\frac{d}{d t} (\frac{ln ( 2 t - 1 )}{t ^{2} - 1})

What is the integral of θ\sqrt[4]{1-θ^2} ?
The integral of θ\sqrt[4]{1-θ^2} is -2/5 (-θ^2+1)^{5/4}+C