What is the integral of sin^{1/2}(2z)cos^3(2z) ?

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

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integral of sin^{1/2}(2z)cos^3(2z)

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

\frac{1}{3} sin^{\frac{3}{2}} (2 z) - \frac{1}{7} sin^{\frac{7}{2}} (2 z) + C

Solution steps

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

Apply u-substitution

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

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - sin^{2} (u)) cos (u) sin (u) d u

Apply u-substitution

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

Expand

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

= \frac{1}{2} \cdot \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} sin^{\frac{3}{2}} (2 z) - \frac{2}{7} sin^{\frac{7}{2}} (2 z))

Simplify

\frac{1}{2} (\frac{2}{3} sin^{\frac{3}{2}} (2 z) - \frac{2}{7} sin^{\frac{7}{2}} (2 z)) : \frac{1}{3} sin^{\frac{3}{2}} (2 z) - \frac{1}{7} sin^{\frac{7}{2}} (2 z)

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

Add a constant to the solution

= \frac{1}{3} sin^{\frac{3}{2}} (2 z) - \frac{1}{7} sin^{\frac{7}{2}} (2 z) + C

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