What is the integral of-sin(x)cos(nx) ?

The integral of-sin(x)cos(nx) is -1/2 (-1/(1+n)cos(x+nx)-1/(1-n)cos(x-nx))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of-sin(x)cos(nx)

\int - sin (x) cos (n x) d x

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

Solution steps

\int - sin (x) cos (n x) d x

Take the constant out:

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

= - \int sin (x) cos (n x) d x

Use the following identity:

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

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

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

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

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

Add a constant to the solution

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

What is the integral of-sin(x)cos(nx) ?
The integral of-sin(x)cos(nx) is -1/2 (-1/(1+n)cos(x+nx)-1/(1-n)cos(x-nx))+C