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

The integral of cos^5(2t)sin^2(2t) is 1/6 sin^3(2t)-1/5 sin^5(2t)+1/14 sin^7(2t)+C

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

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

\frac{1}{6} sin^{3} (2 t) - \frac{1}{5} sin^{5} (2 t) + \frac{1}{14} sin^{7} (2 t) + C

Solution steps

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

Apply u-substitution

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

Rewrite using trig identities

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

Apply u-substitution

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

Expand

v^{2} (1 - v^{2})^{2} : v^{2} - 2 v^{4} + v^{6}

= \frac{1}{2} \cdot \int v^{2} - 2 v^{4} + v^{6} 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^{2} d v - \int 2 v^{4} d v + \int v^{6} d v)

\int v^{2} d v = \frac{v ^{3}}{3}

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{2} (\frac{v ^{3}}{3} - \frac{2 v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= \frac{1}{2} (\frac{sin ^{3} ( 2 t )}{3} - \frac{2 sin ^{5} ( 2 t )}{5} + \frac{sin ^{7} ( 2 t )}{7})

Simplify

\frac{1}{2} (\frac{sin ^{3} ( 2 t )}{3} - \frac{2 sin ^{5} ( 2 t )}{5} + \frac{sin ^{7} ( 2 t )}{7}) : \frac{1}{6} sin^{3} (2 t) - \frac{1}{5} sin^{5} (2 t) + \frac{1}{14} sin^{7} (2 t)

= \frac{1}{6} sin^{3} (2 t) - \frac{1}{5} sin^{5} (2 t) + \frac{1}{14} sin^{7} (2 t)

Add a constant to the solution

= \frac{1}{6} sin^{3} (2 t) - \frac{1}{5} sin^{5} (2 t) + \frac{1}{14} sin^{7} (2 t) + C

What is the integral of cos^5(2t)sin^2(2t) ?
The integral of cos^5(2t)sin^2(2t) is 1/6 sin^3(2t)-1/5 sin^5(2t)+1/14 sin^7(2t)+C