What is the integral of sin^3(4x)cos^5(4x) ?

The integral of sin^3(4x)cos^5(4x) is 1/4 (-(cos^6(4x))/6+(cos^8(4x))/8)+C

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integral of sin^3(4x)cos^5(4x)

\int sin^{3} (4 x) cos^{5} (4 x) d x

\frac{1}{4} (- \frac{cos ^{6} ( 4 x )}{6} + \frac{cos ^{8} ( 4 x )}{8}) + C

Solution steps

\int sin^{3} (4 x) cos^{5} (4 x) d x

Apply u-substitution

: \int sin^{3} (u) cos^{5} (u) \frac{1}{4} d u

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

Rewrite using trig identities

= \frac{1}{4} \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}{4} \cdot \int - v^{5} (1 - v^{2}) d v

Expand

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

= \frac{1}{4} \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}{4} (- \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}{4} (- \frac{v ^{6}}{6} + \frac{v ^{8}}{8})

Substitute back

= \frac{1}{4} (- \frac{cos ^{6} ( 4 x )}{6} + \frac{cos ^{8} ( 4 x )}{8})

Add a constant to the solution

= \frac{1}{4} (- \frac{cos ^{6} ( 4 x )}{6} + \frac{cos ^{8} ( 4 x )}{8}) + C

y = x + e^{x} - \frac{1}{x}

x^{2} + x y + 2 y^{2} = 58, (- 5, - 3)

\int y^{3} ln (y) d y

\frac{d}{d x} (91 + 2 \cdot 4)

\int \frac{2}{x ^{2}} d x

What is the integral of sin^3(4x)cos^5(4x) ?
The integral of sin^3(4x)cos^5(4x) is 1/4 (-(cos^6(4x))/6+(cos^8(4x))/8)+C