What is the integral of sin^3(2x)cos^8(2x) ?

The integral of sin^3(2x)cos^8(2x) is 1/2 (-(cos^9(2x))/9+(cos^{11}(2x))/(11))+C

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integral of sin^3(2x)cos^8(2x)

\int sin^{3} (2 x) cos^{8} (2 x) d x

\frac{1}{2} (- \frac{cos ^{9} ( 2 x )}{9} + \frac{cos ^{11} ( 2 x )}{11}) + C

Solution steps

\int sin^{3} (2 x) cos^{8} (2 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{8} (u) \frac{1}{2} d u

= \int sin^{3} (u) cos^{8} (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 sin^{3} (u) cos^{8} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{8} (u) d u

Apply u-substitution

: \int - v^{8} (1 - v^{2}) d v

= \frac{1}{2} \cdot \int - v^{8} (1 - v^{2}) d v

Expand

- v^{8} (1 - v^{2}) : - v^{8} + v^{10}

= \frac{1}{2} \cdot \int - v^{8} + v^{10} 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^{8} d v + \int v^{10} d v)

\int v^{8} d v = \frac{v ^{9}}{9}

\int v^{10} d v = \frac{v ^{11}}{11}

= \frac{1}{2} (- \frac{v ^{9}}{9} + \frac{v ^{11}}{11})

Substitute back

= \frac{1}{2} (- \frac{cos ^{9} ( 2 x )}{9} + \frac{cos ^{11} ( 2 x )}{11})

Add a constant to the solution

= \frac{1}{2} (- \frac{cos ^{9} ( 2 x )}{9} + \frac{cos ^{11} ( 2 x )}{11}) + C

\frac{\partial}{\partial x} (x^{3} y^{4})

\frac{x - 4}{x + 4}

\frac{\partial}{\partial x} ((x^{2} + y^{2}) ln (x y) + 5 y^{5} e^{3 x^{4} y})

x + 2

x \to - 2 + lim (- x - 2 + 4)

What is the integral of sin^3(2x)cos^8(2x) ?
The integral of sin^3(2x)cos^8(2x) is 1/2 (-(cos^9(2x))/9+(cos^{11}(2x))/(11))+C