What is the integral of sin^4(3x) ?

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

\int sin^{4} (3 x) d x

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

Solution steps

\int sin^{4} (3 x) d x

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 x

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

Simplify

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

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

Add a constant to the solution

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

A e^{- 4}

\int \frac{x}{x ^{2} + x - 6} d x

\frac{2}{3 ( s + 1 ) ^{2} - 27}

\int \frac{x + 5}{x ^{2} + 9} d x

\int \frac{x - 1}{( x - 2 ) ( x - 3 )} d x

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