What is the integral of+cos(x)cos(7x) ?

The integral of+cos(x)cos(7x) is 1/2 (1/6 sin(6x)+1/8 sin(8x))+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(7x)

\int + cos (x) cos (7 x) d x

\frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{8} sin (8 x)) + C

Solution steps

\int cos (x) cos (7 x) d x

Rewrite using trig identities

= \int \frac{1}{2} (cos (8 x) + cos (- 6 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 (8 x) + cos (- 6 x) d x

Simplify

cos (8 x) + cos (- 6 x) : cos (6 x) + cos (8 x)

= \frac{1}{2} \cdot \int cos (6 x) + cos (8 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 (6 x) d x + \int cos (8 x) d x)

\int cos (6 x) d x = \frac{1}{6} sin (6 x)

\int cos (8 x) d x = \frac{1}{8} sin (8 x)

= \frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{8} sin (8 x))

Add a constant to the solution

= \frac{1}{2} (\frac{1}{6} sin (6 x) + \frac{1}{8} sin (8 x)) + C

\int \frac{w - 1}{2 w ^{2} - 2 w + 1} d w

e^{(2 x)}

\int \frac{sin ( h ) 7 x}{1 + cos ( h ^{2} ) 7 x} d x

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

What is the integral of+cos(x)cos(7x) ?
The integral of+cos(x)cos(7x) is 1/2 (1/6 sin(6x)+1/8 sin(8x))+C