What is the integral of sin(9x)cos(7x) ?

The integral of sin(9x)cos(7x) is 1/2 (-1/16 cos(16x)-1/2 cos(2x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(9x)cos(7x)

\int sin (9 x) cos (7 x) d x

\frac{1}{2} (- \frac{1}{16} cos (16 x) - \frac{1}{2} cos (2 x)) + C

Solution steps

\int sin (9 x) cos (7 x) d x

Use the following identity:

cos (t) sin (s) = \frac{sin ( s + t ) + sin ( s - t )}{2}

= \int \frac{sin ( 9 x + 7 x ) + sin ( 9 x - 7 x )}{2} 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 sin (9 x + 7 x) + sin (9 x - 7 x) d x

Simplify

= \frac{1}{2} \cdot \int sin (16 x) + sin (2 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 sin (16 x) d x + \int sin (2 x) d x)

\int sin (16 x) d x = - \frac{1}{16} cos (16 x)

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

= \frac{1}{2} (- \frac{1}{16} cos (16 x) - \frac{1}{2} cos (2 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{16} cos (16 x) - \frac{1}{2} cos (2 x)) + C

\int x cos^{2} (7 x) d x

(sin (5 x))^{^{'}}

f (x) = x + \frac{1}{x}

n = 0 \sum \infty \frac{2}{3 ^{n}}

f (x) = (1 + x^{2}) (1 + x)

What is the integral of sin(9x)cos(7x) ?
The integral of sin(9x)cos(7x) is 1/2 (-1/16 cos(16x)-1/2 cos(2x))+C