What is the integral of sin^3(6θ)sqrt(cos(6θ)) ?

The integral of sin^3(6θ)sqrt(cos(6θ)) is -1/6 (2/3 cos^{3/2}(6θ)-2/7 cos^{7/2}(6θ))+C

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integral of sin^3(6θ)sqrt(cos(6θ))

\int sin^{3} (6 θ) cos (6 θ) d θ

- \frac{1}{6} (\frac{2}{3} cos^{\frac{3}{2}} (6 θ) - \frac{2}{7} cos^{\frac{7}{2}} (6 θ)) + C

Solution steps

\int sin^{3} (6 θ) cos (6 θ) d θ

Apply u-substitution

= \int sin^{3} (u) cos (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sin^{3} (u) cos (u) d u

Apply u-substitution

= \frac{1}{6} \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}{6} (- \int v (1 - v^{2}) d v)

Expand

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

= \frac{1}{6} (- \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}{6} (- (\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}{6} (- (\frac{2}{3} v^{\frac{3}{2}} - \frac{2}{7} v^{\frac{7}{2}}))

Substitute back

= \frac{1}{6} (- (\frac{2}{3} cos^{\frac{3}{2}} (6 θ) - \frac{2}{7} cos^{\frac{7}{2}} (6 θ)))

Simplify

= - \frac{1}{6} (\frac{2}{3} cos^{\frac{3}{2}} (6 θ) - \frac{2}{7} cos^{\frac{7}{2}} (6 θ))

Add a constant to the solution

= - \frac{1}{6} (\frac{2}{3} cos^{\frac{3}{2}} (6 θ) - \frac{2}{7} cos^{\frac{7}{2}} (6 θ)) + C

What is the integral of sin^3(6θ)sqrt(cos(6θ)) ?
The integral of sin^3(6θ)sqrt(cos(6θ)) is -1/6 (2/3 cos^{3/2}(6θ)-2/7 cos^{7/2}(6θ))+C