What is the integral of sin^3(2θ)sqrt(cos(2θ)) ?

The integral of sin^3(2θ)sqrt(cos(2θ)) is -1/3 cos^{3/2}(2θ)+1/7 cos^{7/2}(2θ)+C

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integral of sin^3(2θ)sqrt(cos(2θ))

\int sin^{3} (2 θ) cos (2 θ) d θ

- \frac{1}{3} cos^{\frac{3}{2}} (2 θ) + \frac{1}{7} cos^{\frac{7}{2}} (2 θ) + C

Solution steps

\int sin^{3} (2 θ) cos (2 θ) d θ

Apply u-substitution

: \int sin^{3} (u) cos (u) \frac{1}{2} d u

= \int sin^{3} (u) cos (u) \frac{1}{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 sin^{3} (u) cos (u) d u

Apply u-substitution

: \int - v (1 - v^{2}) d v

= \frac{1}{2} \cdot \int - v (1 - v^{2}) d v

Take the constant out:

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

= \frac{1}{2} (- \int v (1 - v^{2}) d v)

Expand

v (1 - v^{2}) : v - v^{\frac{5}{2}}

= \frac{1}{2} (- \int v - v^{\frac{5}{2}} d v)

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} (- (\int v d v - \int v^{\frac{5}{2}} d v))

\int v d v = \frac{2}{3} v^{\frac{3}{2}}

\int v^{\frac{5}{2}} d v = \frac{2}{7} v^{\frac{7}{2}}

= \frac{1}{2} (- (\frac{2}{3} v^{\frac{3}{2}} - \frac{2}{7} v^{\frac{7}{2}}))

Substitute back

= \frac{1}{2} (- (\frac{2}{3} cos^{\frac{3}{2}} (2 θ) - \frac{2}{7} cos^{\frac{7}{2}} (2 θ)))

Simplify

\frac{1}{2} (- (\frac{2}{3} cos^{\frac{3}{2}} (2 θ) - \frac{2}{7} cos^{\frac{7}{2}} (2 θ))) : - \frac{1}{3} cos^{\frac{3}{2}} (2 θ) + \frac{1}{7} cos^{\frac{7}{2}} (2 θ)

= - \frac{1}{3} cos^{\frac{3}{2}} (2 θ) + \frac{1}{7} cos^{\frac{7}{2}} (2 θ)

Add a constant to the solution

= - \frac{1}{3} cos^{\frac{3}{2}} (2 θ) + \frac{1}{7} cos^{\frac{7}{2}} (2 θ) + C

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What is the integral of sin^3(2θ)sqrt(cos(2θ)) ?
The integral of sin^3(2θ)sqrt(cos(2θ)) is -1/3 cos^{3/2}(2θ)+1/7 cos^{7/2}(2θ)+C