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

The integral of sin^3(2x)cos^4(2x) is 1/2 (-(cos^5(2x))/5+(cos^7(2x))/7)+C

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

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

\frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7}) + C

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int - v^{4} (1 - v^{2}) d v

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

Expand

- v^{4} (1 - v^{2}) : - v^{4} + v^{6}

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

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

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{2} (- \frac{v ^{5}}{5} + \frac{v ^{7}}{7})

Substitute back

= \frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7})

Add a constant to the solution

= \frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5} + \frac{cos ^{7} ( 2 x )}{7}) + C

\frac{\partial}{\partial x} (2 x e^{x} - y + 6 x^{2})

\int \frac{x ^{3} - 3 x ^{2} + 1}{x} d x

\frac{d}{d x} (x i)

\frac{x}{2 - tan ( x )}

3 (x + 1), 4 (x + 1), 0, 3

What is the integral of sin^3(2x)cos^4(2x) ?
The integral of sin^3(2x)cos^4(2x) is 1/2 (-(cos^5(2x))/5+(cos^7(2x))/7)+C