What is the integral of sin(4x)cos(x) ?

The integral of sin(4x)cos(x) is 1/2 (-1/5 cos(5x)-1/3 cos(3x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(4x)cos(x)

\int sin (4 x) cos (x) d x

\frac{1}{2} (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x)) + C

Solution steps

\int sin (4 x) cos (x) d x

Use the following identity:

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

= \int \frac{sin ( 4 x + x ) + sin ( 4 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 (4 x + x) + sin (4 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 (4 x + x) d x + \int sin (4 x - x) d x)

\int sin (4 x + x) d x = - \frac{1}{5} cos (5 x)

\int sin (4 x - x) d x = - \frac{1}{3} cos (3 x)

= \frac{1}{2} (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{1}{2} (- \frac{1}{5} cos (5 x) - \frac{1}{3} cos (3 x)) + C

\int \frac{x ^{2} + x + a}{x ^{3} - 9 x ^{2} + 27 x - 27} d x

\int \frac{sec ^{2} ( 6 x )}{3 tan ^{2} ( 6 x ) + 15} d x

\int \frac{1}{x ^{2} - 4 x - 5} d x

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

\int \frac{4 x ^{2} - 49}{x} d x

What is the integral of sin(4x)cos(x) ?
The integral of sin(4x)cos(x) is 1/2 (-1/5 cos(5x)-1/3 cos(3x))+C