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

The integral of sin(2x)cos^5(2x) is -1/12 cos^6(2x)+C

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

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

- \frac{1}{12} cos^{6} (2 x) + C

Solution steps

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

Apply u-substitution

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

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

Apply u-substitution

: \int - v^{5} d v

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

Apply the Power Rule

: \frac{v ^{6}}{6}

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

Substitute back

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

Simplify

\frac{1}{2} (- \frac{cos ^{6} ( 2 x )}{6}) : - \frac{1}{12} cos^{6} (2 x)

= - \frac{1}{12} cos^{6} (2 x)

Add a constant to the solution

= - \frac{1}{12} cos^{6} (2 x) + C

\frac{\partial}{\partial x} (x^{4} sin (y^{9} z^{9}))

x \to 5 lim (\frac{x ^{2} - 25}{3 x - 15})

(1 - x)^{2} + x^{3}

x \to 5 lim (\frac{1}{x ^{2} + x})

What is the integral of sin(2x)cos^5(2x) ?
The integral of sin(2x)cos^5(2x) is -1/12 cos^6(2x)+C