What is the integral of sin^4(2θ) ?

The integral of sin^4(2θ) is 1/2 (-1/4 sin^3(2θ)cos(2θ)+3/8 (2θ-1/2 sin(4θ)))+C

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integral of sin^4(2θ)

\int sin^{4} (2 θ) d θ

\frac{1}{2} (- \frac{1}{4} sin^{3} (2 θ) cos (2 θ) + \frac{3}{8} (2 θ - \frac{1}{2} sin (4 θ))) + C

Solution steps

\int sin^{4} (2 θ) d θ

Apply u-substitution

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

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

Apply Integral Reduction

= \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

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

= \frac{1}{2} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 2 θ

= \frac{1}{2} (- \frac{cos ( 2 θ ) sin ^{3} ( 2 θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (2 θ - \frac{1}{2} sin (2 \cdot 2 θ)))

Simplify

\frac{1}{2} (- \frac{cos ( 2 θ ) sin ^{3} ( 2 θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (2 θ - \frac{1}{2} sin (2 \cdot 2 θ))) : \frac{1}{2} (- \frac{1}{4} sin^{3} (2 θ) cos (2 θ) + \frac{3}{8} (2 θ - \frac{1}{2} sin (4 θ)))

= \frac{1}{2} (- \frac{1}{4} sin^{3} (2 θ) cos (2 θ) + \frac{3}{8} (2 θ - \frac{1}{2} sin (4 θ)))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{4} sin^{3} (2 θ) cos (2 θ) + \frac{3}{8} (2 θ - \frac{1}{2} sin (4 θ))) + C

\frac{\partial}{\partial y} (x sin (y - z))

x \to 2 lim ((f (x))^{3})

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

\frac{d}{d x} (3 sin (x) - x)

\frac{\partial}{\partial x} (3 x^{2} + 3 y^{2})

What is the integral of sin^4(2θ) ?
The integral of sin^4(2θ) is 1/2 (-1/4 sin^3(2θ)cos(2θ)+3/8 (2θ-1/2 sin(4θ)))+C