What is the integral of 7sin(8x)cos(5x) ?

The integral of 7sin(8x)cos(5x) is 7/2 (-1/13 cos(13x)-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 7sin(8x)cos(5x)

\int 7 sin (8 x) cos (5 x) d x

\frac{7}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x)) + C

Solution steps

\int 7 sin (8 x) cos (5 x) d x

Take the constant out:

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

= 7 \cdot \int sin (8 x) cos (5 x) d x

Use the following identity:

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

= 7 \cdot \int \frac{sin ( 8 x + 5 x ) + sin ( 8 x - 5 x )}{2} d x

Take the constant out:

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

= 7 \cdot \frac{1}{2} \cdot \int sin (8 x + 5 x) + sin (8 x - 5 x) d x

Simplify

= 7 \cdot \frac{1}{2} \cdot \int sin (13 x) + sin (3 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

= 7 \cdot \frac{1}{2} (\int sin (13 x) d x + \int sin (3 x) d x)

\int sin (13 x) d x = - \frac{1}{13} cos (13 x)

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

= 7 \cdot \frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x))

Simplify

7 \cdot \frac{1}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x)) : \frac{7}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x))

= \frac{7}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x))

Add a constant to the solution

= \frac{7}{2} (- \frac{1}{13} cos (13 x) - \frac{1}{3} cos (3 x)) + C

\frac{2}{s} - \frac{1}{s ^{3}}

sin (π x), 0.5

y = x^{c o s (x)}

x \to \infty lim (2 x - 5)

\int 37 arctan (x) d x

What is the integral of 7sin(8x)cos(5x) ?
The integral of 7sin(8x)cos(5x) is 7/2 (-1/13 cos(13x)-1/3 cos(3x))+C