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

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

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integral of sin(2θ)cos^4(2θ)

\int sin (2 θ) cos^{4} (2 θ) d θ

- \frac{1}{10} cos^{5} (2 θ) + C

Solution steps

\int sin (2 θ) cos^{4} (2 θ) d θ

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 θ )}{5})

Simplify

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

= - \frac{1}{10} cos^{5} (2 θ)

Add a constant to the solution

= - \frac{1}{10} cos^{5} (2 θ) + C

\int \frac{1}{sec ( θ )} d θ

\frac{d}{d t} (t cos (t))

cos (5 t) + sin (2 t)

n = 0 \sum \infty \frac{9}{10 0 ^{n}}

\frac{d y}{d x} = \frac{y}{x} ln (\frac{y}{x})

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