What is the integral of sin^2(6x)cos^2(6x) ?

The integral of sin^2(6x)cos^2(6x) is 1/8 (x-1/24 sin(24x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin^2(6x)cos^2(6x)

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

\frac{1}{8} (x - \frac{1}{24} sin (24 x)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{8} (1 - cos (24 x)) d x

Take the constant out:

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

= \frac{1}{8} \cdot \int 1 - cos (24 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int 1 d x - \int cos (24 x) d x)

\int 1 d x = x

\int cos (24 x) d x = \frac{1}{24} sin (24 x)

= \frac{1}{8} (x - \frac{1}{24} sin (24 x))

Add a constant to the solution

= \frac{1}{8} (x - \frac{1}{24} sin (24 x)) + C

x^{2}, 12 - x

\frac{d}{d x} (\frac{x ^{3} - y ^{3}}{x - y})

x \to 0 lim (+ (x^{x}))

y^{^{'}} = 7 t y^{2}

\frac{ln ( x )}{x ^{5}}

What is the integral of sin^2(6x)cos^2(6x) ?
The integral of sin^2(6x)cos^2(6x) is 1/8 (x-1/24 sin(24x))+C