What is the integral of sin^4(6θ) ?

The integral of sin^4(6θ) is 1/6 (-1/4 sin^3(6θ)cos(6θ)+3/8 (6θ-1/2 sin(12θ)))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^4(6θ)

\int sin^{4} (6 θ) d θ

\frac{1}{6} (- \frac{1}{4} sin^{3} (6 θ) cos (6 θ) + \frac{3}{8} (6 θ - \frac{1}{2} sin (12 θ))) + C

Solution steps

\int sin^{4} (6 θ) d θ

Apply u-substitution

: \int sin^{4} (u) \frac{1}{6} d u

= \int sin^{4} (u) \frac{1}{6} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{6} (- \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}{6} (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = 6 θ

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

Simplify

\frac{1}{6} (- \frac{cos ( 6 θ ) sin ^{3} ( 6 θ )}{4} + \frac{3}{4} \cdot \frac{1}{2} (6 θ - \frac{1}{2} sin (2 \cdot 6 θ))) : \frac{1}{6} (- \frac{1}{4} sin^{3} (6 θ) cos (6 θ) + \frac{3}{8} (6 θ - \frac{1}{2} sin (12 θ)))

= \frac{1}{6} (- \frac{1}{4} sin^{3} (6 θ) cos (6 θ) + \frac{3}{8} (6 θ - \frac{1}{2} sin (12 θ)))

Add a constant to the solution

= \frac{1}{6} (- \frac{1}{4} sin^{3} (6 θ) cos (6 θ) + \frac{3}{8} (6 θ - \frac{1}{2} sin (12 θ))) + C

x y^{^{'}} + 2 y = - 3 e^{x}

\int \frac{( ln ( x ) ) ^{20}}{x} d x

\frac{d}{d x} (arccos (5 x))

\frac{\partial}{\partial x} (4 - x^{2} - 2 y)

\int (\frac{1}{4 x ^{2}}) d x

What is the integral of sin^4(6θ) ?
The integral of sin^4(6θ) is 1/6 (-1/4 sin^3(6θ)cos(6θ)+3/8 (6θ-1/2 sin(12θ)))+C