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

The integral of sin(2x)cos^4(2x) is -1/10 cos^5(2x)+C

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

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

- \frac{1}{10} cos^{5} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - v^{4} d v

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

Take the constant out:

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

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

Apply the Power Rule

: \frac{v ^{5}}{5}

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

Substitute back

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

Simplify

\frac{1}{2} (- \frac{cos ^{5} ( 2 x )}{5}) : - \frac{1}{10} cos^{5} (2 x)

= - \frac{1}{10} cos^{5} (2 x)

Add a constant to the solution

= - \frac{1}{10} cos^{5} (2 x) + C

f (x) = (7 x + 6)^{2}

(- 16 x^{2} + 1)^{2}

y^{^{''}} = sin (x)

x (\frac{d y}{d x}) - y = (\frac{x ^{3}}{y}) e^{\frac{y}{x}}

x \to 0 lim (\frac{e ^{∣ x ∣} - 1}{x})

What is the integral of sin(2x)cos^4(2x) ?
The integral of sin(2x)cos^4(2x) is -1/10 cos^5(2x)+C