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

The integral of sin(6x)cos(x) is 1/2 (-1/7 cos(7x)-1/5 cos(5x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(6x)cos(x)

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

\frac{1}{2} (- \frac{1}{7} cos (7 x) - \frac{1}{5} cos (5 x)) + C

Solution steps

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

Use the following identity:

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

= \int \frac{sin ( 6 x + x ) + sin ( 6 x - 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 (6 x + x) + sin (6 x - 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 (6 x + x) d x + \int sin (6 x - x) d x)

\int sin (6 x + x) d x = - \frac{1}{7} cos (7 x)

\int sin (6 x - x) d x = - \frac{1}{5} cos (5 x)

= \frac{1}{2} (- \frac{1}{7} cos (7 x) - \frac{1}{5} cos (5 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{7} cos (7 x) - \frac{1}{5} cos (5 x)) + C

\int \frac{3}{2} (1 - x^{2}) d x

\int (4 x - 5)^{3} d x

\int \frac{1}{64 sec ^{2} ( x ) tan ( x )} d x

\int \frac{1}{9 - 9 x ^{2}} d x

\int x^{9} ln (3 x) d x

What is the integral of sin(6x)cos(x) ?
The integral of sin(6x)cos(x) is 1/2 (-1/7 cos(7x)-1/5 cos(5x))+C