What is the integral of sin^5(2t)cos(2t) ?

The integral of sin^5(2t)cos(2t) is 1/12 sin^6(2t)+C

integral of sin^5(2t)cos(2t)

\int sin^{5} (2 t) cos (2 t) d t

\frac{1}{12} sin^{6} (2 t) + C

Solution steps

\int sin^{5} (2 t) cos (2 t) d t

Apply u-substitution

= \int \frac{5 u}{10} d u

Take the constant out:

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

= \frac{1}{10} \cdot \int 5 u d u

Apply the Power Rule

= \frac{1}{10} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = sin^{5} (2 t)

= \frac{1}{10} \cdot \frac{5}{6} (sin^{5} (2 t))^{\frac{6}{5}}

Simplify

\frac{1}{10} \cdot \frac{5}{6} (sin^{5} (2 t))^{\frac{6}{5}} : \frac{1}{12} sin^{6} (2 t)

= \frac{1}{12} sin^{6} (2 t)

Add a constant to the solution

= \frac{1}{12} sin^{6} (2 t) + C

a re a 3 x^{3} - 2 x, - 3, 3

\frac{\partial}{\partial y} (\frac{1}{t ( y ^{2} - 9 )})

a re a f (x) = \frac{x}{9 - x ^{2}}, g (y) = 0, y = 2

\frac{d}{d x} (arcsin (2 x - 1))

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 e^{2 t}, y (0) = 1, y^{^{'}} (0) = 0

What is the integral of sin^5(2t)cos(2t) ?
The integral of sin^5(2t)cos(2t) is 1/12 sin^6(2t)+C