What is the integral of sin^3(2θ)cot(2θ) ?

The integral of sin^3(2θ)cot(2θ) is 1/12 (sin(6θ)-3sin(2θ)cos(4θ))+C

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integral of sin^3(2θ)cot(2θ)

\int sin^{3} (2 θ) cot (2 θ) d θ

\frac{1}{12} (sin (6 θ) - 3 sin (2 θ) cos (4 θ)) + C

Solution steps

\int sin^{3} (2 θ) cot (2 θ) d θ

Rewrite using trig identities

= \int \frac{1}{2} cos (2 θ) (1 - cos (4 θ)) d θ

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (2 θ) (1 - cos (4 θ)) d θ

Apply Integration By Parts

: \frac{1}{2} sin (2 θ) (1 - cos (4 θ)) - \int 2 sin (4 θ) sin (2 θ) d θ

= \frac{1}{2} (\frac{1}{2} sin (2 θ) (1 - cos (4 θ)) - \int 2 sin (4 θ) sin (2 θ) d θ)

\int 2 sin (4 θ) sin (2 θ) d θ = - \frac{1}{6} sin (6 θ) + \frac{1}{2} sin (2 θ)

= \frac{1}{2} (\frac{1}{2} sin (2 θ) (1 - cos (4 θ)) - (- \frac{1}{6} sin (6 θ) + \frac{1}{2} sin (2 θ)))

Simplify

\frac{1}{2} (\frac{1}{2} sin (2 θ) (1 - cos (4 θ)) - (- \frac{1}{6} sin (6 θ) + \frac{1}{2} sin (2 θ))) : \frac{1}{12} (sin (6 θ) - 3 sin (2 θ) cos (4 θ))

= \frac{1}{12} (sin (6 θ) - 3 sin (2 θ) cos (4 θ))

Add a constant to the solution

= \frac{1}{12} (sin (6 θ) - 3 sin (2 θ) cos (4 θ)) + C

f (x) = 2 x^{3} - x^{2} + 2, (1, 3)

x \to - 1 + lim (f (x))

\int 5 x cos (6 x) d x

x \to 0 lim ((1 + \frac{1}{x})^{3 x})

What is the integral of sin^3(2θ)cot(2θ) ?
The integral of sin^3(2θ)cot(2θ) is 1/12 (sin(6θ)-3sin(2θ)cos(4θ))+C