What is the integral of sin^3(4x)cos^9(4x) ?

The integral of sin^3(4x)cos^9(4x) is 1/4 (-(cos^{10}(4x))/(10)+(cos^{12}(4x))/(12))+C

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integral of sin^3(4x)cos^9(4x)

\int sin^{3} (4 x) cos^{9} (4 x) d x

\frac{1}{4} (- \frac{cos ^{10} ( 4 x )}{10} + \frac{cos ^{12} ( 4 x )}{12}) + C

Solution steps

\int sin^{3} (4 x) cos^{9} (4 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{9} (u) \frac{1}{4} d u

= \int sin^{3} (u) cos^{9} (u) \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int sin^{3} (u) cos^{9} (u) d u

Rewrite using trig identities

= \frac{1}{4} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{9} (u) d u

Apply u-substitution

: \int - v^{9} (1 - v^{2}) d v

= \frac{1}{4} \cdot \int - v^{9} (1 - v^{2}) d v

Expand

- v^{9} (1 - v^{2}) : - v^{9} + v^{11}

= \frac{1}{4} \cdot \int - v^{9} + v^{11} 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}{4} (- \int v^{9} d v + \int v^{11} d v)

\int v^{9} d v = \frac{v ^{10}}{10}

\int v^{11} d v = \frac{v ^{12}}{12}

= \frac{1}{4} (- \frac{v ^{10}}{10} + \frac{v ^{12}}{12})

Substitute back

= \frac{1}{4} (- \frac{cos ^{10} ( 4 x )}{10} + \frac{cos ^{12} ( 4 x )}{12})

Add a constant to the solution

= \frac{1}{4} (- \frac{cos ^{10} ( 4 x )}{10} + \frac{cos ^{12} ( 4 x )}{12}) + C

\int \frac{ln ( x ^{26} )}{x} d x

\frac{\partial}{\partial x} (240 x^{2} - 140 y^{2})

t cos (5 t)

x \to - 9 lim (- 9)

\frac{d}{d v} (uv)

What is the integral of sin^3(4x)cos^9(4x) ?
The integral of sin^3(4x)cos^9(4x) is 1/4 (-(cos^{10}(4x))/(10)+(cos^{12}(4x))/(12))+C