What is the integral of 4sin^3(2x)cos^2(2x) ?

The integral of 4sin^3(2x)cos^2(2x) is 2(-1/3 cos^3(2x)+1/5 cos^5(2x))+C

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integral of 4sin^3(2x)cos^2(2x)

\int 4 sin^{3} (2 x) cos^{2} (2 x) d x

2 (- \frac{1}{3} cos^{3} (2 x) + \frac{1}{5} cos^{5} (2 x)) + C

Solution steps

\int 4 sin^{3} (2 x) cos^{2} (2 x) d x

Take the constant out:

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

= 4 \cdot \int sin^{3} (2 x) cos^{2} (2 x) d x

Apply u-substitution

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

= 4 \cdot \int sin^{3} (u) cos^{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

= 4 \cdot \frac{1}{2} \cdot \int sin^{3} (u) cos^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{2} (u) d u

Apply u-substitution

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

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

Expand

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

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

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

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

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

= 4 \cdot \frac{1}{2} (- \frac{v ^{3}}{3} + \frac{v ^{5}}{5})

Substitute back

= 4 \cdot \frac{1}{2} (- \frac{cos ^{3} ( 2 x )}{3} + \frac{cos ^{5} ( 2 x )}{5})

Simplify

4 \cdot \frac{1}{2} (- \frac{cos ^{3} ( 2 x )}{3} + \frac{cos ^{5} ( 2 x )}{5}) : 2 (- \frac{1}{3} cos^{3} (2 x) + \frac{1}{5} cos^{5} (2 x))

= 2 (- \frac{1}{3} cos^{3} (2 x) + \frac{1}{5} cos^{5} (2 x))

Add a constant to the solution

= 2 (- \frac{1}{3} cos^{3} (2 x) + \frac{1}{5} cos^{5} (2 x)) + C

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What is the integral of 4sin^3(2x)cos^2(2x) ?
The integral of 4sin^3(2x)cos^2(2x) is 2(-1/3 cos^3(2x)+1/5 cos^5(2x))+C