What is the integral of 12sin^2(2x)cos(2x) ?

The integral of 12sin^2(2x)cos(2x) is -3sin(2x)cos(4x)+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 12sin^2(2x)cos(2x)

\int 12 sin^{2} (2 x) cos (2 x) d x

- 3 sin (2 x) cos (4 x) + sin (6 x) + C

Solution steps

\int 12 sin^{2} (2 x) cos (2 x) d x

Take the constant out:

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

= 12 \cdot \int sin^{2} (2 x) cos (2 x) d x

Rewrite using trig identities

= 12 \cdot \int \frac{1}{2} cos (2 x) (1 - cos (4 x)) d x

Take the constant out:

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

= 12 \cdot \frac{1}{2} \cdot \int cos (2 x) (1 - cos (4 x)) d x

Apply Integration By Parts

: \frac{1}{2} sin (2 x) (1 - cos (4 x)) - \int 2 sin (4 x) sin (2 x) d x

= 12 \cdot \frac{1}{2} (\frac{1}{2} sin (2 x) (1 - cos (4 x)) - \int 2 sin (4 x) sin (2 x) d x)

\int 2 sin (4 x) sin (2 x) d x = - \frac{1}{6} sin (6 x) + \frac{1}{2} sin (2 x)

= 12 \cdot \frac{1}{2} (\frac{1}{2} sin (2 x) (1 - cos (4 x)) - (- \frac{1}{6} sin (6 x) + \frac{1}{2} sin (2 x)))

Simplify

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

= - 3 sin (2 x) cos (4 x) + sin (6 x)

Add a constant to the solution

= - 3 sin (2 x) cos (4 x) + sin (6 x) + C

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

\int \frac{3 - x}{2 x ^{2} - 12 x + 1} d x

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

\int sin (6 x) sin (7 x) d x

\int \frac{2 x ^{2} cos ( 4 x ) - 3}{x ^{2}} d x

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