What is the integral of (cos(3x))(1sin(3x)-1) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (cos(3x))(1sin(3x)-1)

\int (cos (3 x)) (1 sin (3 x) - 1) d x

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

Solution steps

\int cos (3 x) (1 \cdot sin (3 x) - 1) d x

Expand

cos (3 x) (1 \cdot sin (3 x) - 1) : cos (3 x) sin (3 x) - cos (3 x)

= \int cos (3 x) sin (3 x) - cos (3 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

= \int cos (3 x) sin (3 x) d x - \int cos (3 x) d x

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

\int cos (3 x) d x = \frac{1}{3} sin (3 x)

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

Add a constant to the solution

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

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