What is the integral of sin^4(3θ) ?

The integral of sin^4(3θ) is -1/12 sin^3(3θ)cos(3θ)+1/16 (6θ-sin(6θ))+C

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integral of sin^4(3θ)

\int sin^{4} (3 θ) d θ

- \frac{1}{12} sin^{3} (3 θ) cos (3 θ) + \frac{1}{16} (6 θ - sin (6 θ)) + C

Solution steps

\int sin^{4} (3 θ) d θ

Apply u-substitution

: \int sin^{4} (u) \frac{1}{3} d u

= \int sin^{4} (u) \frac{1}{3} d u

Take the constant out:

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

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

Apply Integral Reduction

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

Substitute back

u = 3 θ

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

Simplify

\frac{1}{3} (- \frac{cos ( 3 θ ) sin ^{3} ( 3 θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (3 θ - \frac{1}{2} sin (2 \cdot 3 θ))) : - \frac{1}{12} sin^{3} (3 θ) cos (3 θ) + \frac{1}{16} (6 θ - sin (6 θ))

= - \frac{1}{12} sin^{3} (3 θ) cos (3 θ) + \frac{1}{16} (6 θ - sin (6 θ))

Add a constant to the solution

= - \frac{1}{12} sin^{3} (3 θ) cos (3 θ) + \frac{1}{16} (6 θ - sin (6 θ)) + C

\int \frac{4}{x ln ( x )} d x

\int_{0}^{5} (4 x - x^{2}) d x

\int 6 sin^{2} (x) d x

\frac{d}{d x} (11 sec^{2} (x))

What is the integral of sin^4(3θ) ?
The integral of sin^4(3θ) is -1/12 sin^3(3θ)cos(3θ)+1/16 (6θ-sin(6θ))+C