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

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

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integral of sqrt(cos(3x))sin^3(3x)

\int cos (3 x) sin^{3} (3 x) d x

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

Solution steps

\int cos (3 x) sin^{3} (3 x) d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

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

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

Expand

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

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

Substitute back

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

Simplify

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

Add a constant to the solution

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

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