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

The integral of 2sin^3(2x)cos^3(2x) is -1/4 cos^4(2x)+1/6 cos^6(2x)+C

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

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

- \frac{1}{4} cos^{4} (2 x) + \frac{1}{6} cos^{6} (2 x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 2 \cdot \int sin^{3} (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

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

Rewrite using trig identities

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

Apply u-substitution

: \int - v^{3} (1 - v^{2}) d v

= 2 \cdot \frac{1}{2} \cdot \int - v^{3} (1 - v^{2}) d v

Expand

- v^{3} (1 - v^{2}) : - v^{3} + v^{5}

= 2 \cdot \frac{1}{2} \cdot \int - v^{3} + v^{5} 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

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

\int v^{3} d v = \frac{v ^{4}}{4}

\int v^{5} d v = \frac{v ^{6}}{6}

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

Substitute back

= 2 \cdot \frac{1}{2} (- \frac{cos ^{4} ( 2 x )}{4} + \frac{cos ^{6} ( 2 x )}{6})

Simplify

2 \cdot \frac{1}{2} (- \frac{cos ^{4} ( 2 x )}{4} + \frac{cos ^{6} ( 2 x )}{6}) : - \frac{1}{4} cos^{4} (2 x) + \frac{1}{6} cos^{6} (2 x)

= - \frac{1}{4} cos^{4} (2 x) + \frac{1}{6} cos^{6} (2 x)

Add a constant to the solution

= - \frac{1}{4} cos^{4} (2 x) + \frac{1}{6} cos^{6} (2 x) + C

x \to + 10 lim (x)

\frac{d y}{d x} = x^{6} y^{- 7}, y (0) = 4

x \frac{d y}{d x} - (1 - x) y = x y^{2}

f (x) = e^{- x} - e^{- x} x

\frac{d}{d x} (arcsin (cos (x)))

What is the integral of 2sin^3(2x)cos^3(2x) ?
The integral of 2sin^3(2x)cos^3(2x) is -1/4 cos^4(2x)+1/6 cos^6(2x)+C