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

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

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integral of sin^3(2x)sqrt(cos(2x))

\int sin^{3} (2 x) cos (2 x) d x

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

Solution steps

\int sin^{3} (2 x) cos (2 x) d x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} ((4 x^{2} - \frac{1}{2} x) (9 x + 8))

\frac{d}{d x} (3 e^{4 x})

2 y^{^{'}} = 3 x y + 1

x \to 1 lim ((x^{2} - 2 x + 1) cos (e^{\frac{1}{x - 1}}))

x \to \infty lim ([2, x])

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