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

The integral of (sin^4(3x)-cos^4(3x)) is xsin^4(3x)-xcos^4(3x)+2xcos^2(3x)-x-1/6 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)-cos^4(3x))

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

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

Solution steps

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

Apply Integration By Parts

: (sin^{4} (3 x) - cos^{4} (3 x)) x - \int 12 cos (3 x) sin (3 x) x d x

= (sin^{4} (3 x) - cos^{4} (3 x)) x - \int 12 cos (3 x) sin (3 x) x d x

\int 12 cos (3 x) sin (3 x) x d x = - 2 x cos^{2} (3 x) + x + \frac{1}{6} sin (6 x)

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

Simplify

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

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

Add a constant to the solution

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

\int r^{3} 1 + 4 r^{2} d r

\frac{1}{3 ( 1 + e ^{- t} ) ^{3}}

x d y + 3 y d x = x y d y

x \to + 0 lim (\frac{- 6}{x ^{7}})

What is the integral of (sin^4(3x)-cos^4(3x)) ?
The integral of (sin^4(3x)-cos^4(3x)) is xsin^4(3x)-xcos^4(3x)+2xcos^2(3x)-x-1/6 sin(6x)+C