What is the integral of sin^3(2x)cos^5(2x) ?

The integral of sin^3(2x)cos^5(2x) is 1/2 (-(cos^6(2x))/6+(cos^8(2x))/8)+C

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integral of sin^3(2x)cos^5(2x)

\int sin^{3} (2 x) cos^{5} (2 x) d x

\frac{1}{2} (- \frac{cos ^{6} ( 2 x )}{6} + \frac{cos ^{8} ( 2 x )}{8}) + C

Solution steps

\int sin^{3} (2 x) cos^{5} (2 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{5} (u) \frac{1}{2} d u

= \int sin^{3} (u) cos^{5} (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^{5} (u) d u

Rewrite using trig identities

= \frac{1}{2} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{5} (u) d u

Apply u-substitution

: \int - v^{5} (1 - v^{2}) d v

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

Expand

- v^{5} (1 - v^{2}) : - v^{5} + v^{7}

= \frac{1}{2} \cdot \int - v^{5} + v^{7} 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^{5} d v + \int v^{7} d v)

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

\int v^{7} d v = \frac{v ^{8}}{8}

= \frac{1}{2} (- \frac{v ^{6}}{6} + \frac{v ^{8}}{8})

Substitute back

= \frac{1}{2} (- \frac{cos ^{6} ( 2 x )}{6} + \frac{cos ^{8} ( 2 x )}{8})

Add a constant to the solution

= \frac{1}{2} (- \frac{cos ^{6} ( 2 x )}{6} + \frac{cos ^{8} ( 2 x )}{8}) + C

\int_{0}^{1} x^{2} 3 x^{2} d x

\frac{d}{d x} (ln (7 x^{2} + 3 y^{2}))

\frac{d}{d x} (\frac{x + 1}{x ^{5}})

\frac{\partial}{\partial y} (4 x^{2} y - 5 x y^{2})

y = 9 - 0.25 x^{2}, y = 0, [2, 5]

What is the integral of sin^3(2x)cos^5(2x) ?
The integral of sin^3(2x)cos^5(2x) is 1/2 (-(cos^6(2x))/6+(cos^8(2x))/8)+C