What is the integral of cos(6x)cos(9x) ?

The integral of cos(6x)cos(9x) is 1/2 (1/3 sin(3x)+1/15 sin(15x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cos(6x)cos(9x)

\int cos (6 x) cos (9 x) d x

\frac{1}{2} (\frac{1}{3} sin (3 x) + \frac{1}{15} sin (15 x)) + C

Solution steps

\int cos (6 x) cos (9 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (15 x) + cos (- 3 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (15 x) + cos (- 3 x) d x

Simplify

cos (15 x) + cos (- 3 x) : cos (3 x) + cos (15 x)

= \frac{1}{2} \cdot \int cos (3 x) + cos (15 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}{2} (\int cos (3 x) d x + \int cos (15 x) d x)

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

\int cos (15 x) d x = \frac{1}{15} sin (15 x)

= \frac{1}{2} (\frac{1}{3} sin (3 x) + \frac{1}{15} sin (15 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{3} sin (3 x) + \frac{1}{15} sin (15 x)) + C

\int t ln (t^{2}) d t

\int (x^{2} + 2 x) x^{3} + 3 x^{2} - 5 d x

\int \frac{1}{( 16 + x ^{2} ) ^{4}} d x

\int \frac{x - 4}{x ^{3} + x ^{2} - 2 x} d x

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

What is the integral of cos(6x)cos(9x) ?
The integral of cos(6x)cos(9x) is 1/2 (1/3 sin(3x)+1/15 sin(15x))+C