What is the integral of cos(x)cos(3x) ?

The integral of cos(x)cos(3x) is 1/2 (1/2 sin(2x)+1/4 sin(4x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of cos(x)cos(3x)

\int cos (x) cos (3 x) d x

\frac{1}{2} (\frac{1}{2} sin (2 x) + \frac{1}{4} sin (4 x)) + C

Solution steps

\int cos (x) cos (3 x) d x

Rewrite using trig identities

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

Simplify

cos (4 x) + cos (- 2 x) : cos (2 x) + cos (4 x)

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

\int cos (2 x) d x = \frac{1}{2} sin (2 x)

\int cos (4 x) d x = \frac{1}{4} sin (4 x)

= \frac{1}{2} (\frac{1}{2} sin (2 x) + \frac{1}{4} sin (4 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{2} sin (2 x) + \frac{1}{4} sin (4 x)) + C

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

\frac{6 t}{6 + t}

\frac{d y}{d x} = x y + 9 x + 2 y + 18

36 y^{^{''}} - 60 y^{^{'}} + 26 y = 0, y (4) = 2, y^{^{'}} (4) = 5

\frac{( x + 2 ) ( x ^{2} - 2 x + 4 )}{x ^{3}}

What is the integral of cos(x)cos(3x) ?
The integral of cos(x)cos(3x) is 1/2 (1/2 sin(2x)+1/4 sin(4x))+C