What is the integral of 6sin^3(4x)cos^3(4x) ?

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

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integral of 6sin^3(4x)cos^3(4x)

\int 6 sin^{3} (4 x) cos^{3} (4 x) d x

\frac{3}{2} (- \frac{1}{4} cos^{4} (4 x) + \frac{1}{6} cos^{6} (4 x)) + C

Solution steps

\int 6 sin^{3} (4 x) cos^{3} (4 x) d x

Take the constant out:

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

= 6 \cdot \int sin^{3} (4 x) cos^{3} (4 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{3} (u) \frac{1}{4} d u

= 6 \cdot \int sin^{3} (u) cos^{3} (u) \frac{1}{4} d u

Take the constant out:

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

= 6 \cdot \frac{1}{4} \cdot \int sin^{3} (u) cos^{3} (u) d u

Rewrite using trig identities

= 6 \cdot \frac{1}{4} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{3} (u) d u

Apply u-substitution

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

= 6 \cdot \frac{1}{4} \cdot \int - v^{3} (1 - v^{2}) d v

Expand

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

= 6 \cdot \frac{1}{4} \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

= 6 \cdot \frac{1}{4} (- \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}

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

Substitute back

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

Simplify

6 \cdot \frac{1}{4} (- \frac{cos ^{4} ( 4 x )}{4} + \frac{cos ^{6} ( 4 x )}{6}) : \frac{3}{2} (- \frac{1}{4} cos^{4} (4 x) + \frac{1}{6} cos^{6} (4 x))

= \frac{3}{2} (- \frac{1}{4} cos^{4} (4 x) + \frac{1}{6} cos^{6} (4 x))

Add a constant to the solution

= \frac{3}{2} (- \frac{1}{4} cos^{4} (4 x) + \frac{1}{6} cos^{6} (4 x)) + C

\frac{d y}{d x} = \frac{x ^{3} - 2 y}{x}

\int \frac{( 3 x + 2 )}{19 - 5 x + x ^{2}} d x

x^{3} - 4 x^{2}

y^{^{'}} = - a y^{2} + b

\frac{d}{d x} (tan (1 + e^{2 x}))

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