What is the integral of 9cos(4x)cos(9x) ?

The integral of 9cos(4x)cos(9x) is 9/2 (1/5 sin(5x)+1/13 sin(13x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 9cos(4x)cos(9x)

\int 9 cos (4 x) cos (9 x) d x

\frac{9}{2} (\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x)) + C

Solution steps

\int 9 cos (4 x) cos (9 x) d x

Take the constant out:

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

= 9 \cdot \int cos (4 x) cos (9 x) d x

Rewrite using trig identities

= 9 \cdot \int \frac{1}{2} (cos (13 x) + cos (- 5 x)) d x

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int cos (13 x) + cos (- 5 x) d x

Simplify

cos (13 x) + cos (- 5 x) : cos (5 x) + cos (13 x)

= 9 \cdot \frac{1}{2} \cdot \int cos (5 x) + cos (13 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

= 9 \cdot \frac{1}{2} (\int cos (5 x) d x + \int cos (13 x) d x)

\int cos (5 x) d x = \frac{1}{5} sin (5 x)

\int cos (13 x) d x = \frac{1}{13} sin (13 x)

= 9 \cdot \frac{1}{2} (\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x))

Simplify

9 \cdot \frac{1}{2} (\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x)) : \frac{9}{2} (\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x))

= \frac{9}{2} (\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x))

Add a constant to the solution

= \frac{9}{2} (\frac{1}{5} sin (5 x) + \frac{1}{13} sin (13 x)) + C

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

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

\int - 10 csc^{2} (5 x) cot^{3} (5 x) d x

\int x 16 - x d x

\int \frac{3 x}{- 2} d x

What is the integral of 9cos(4x)cos(9x) ?
The integral of 9cos(4x)cos(9x) is 9/2 (1/5 sin(5x)+1/13 sin(13x))+C