What is the integral of 13cos(pix)cos(4pix) ?

The integral of 13cos(pix)cos(4pix) is 13/2 (1/(3pi)sin(3pix)+1/(5pi)sin(5pix))+C

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integral of 13cos(pix)cos(4pix)

\int 13 cos (π x) cos (4 π x) d x

\frac{13}{2} (\frac{1}{3 π} sin (3 π x) + \frac{1}{5 π} sin (5 π x)) + C

Solution steps

\int 13 cos (π x) cos (4 π x) d x

Take the constant out:

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

= 13 \cdot \int cos (π x) cos (4 π x) d x

Rewrite using trig identities

= 13 \cdot \int \frac{1}{2} (cos (5 π x) + cos (- 3 π x)) d x

Take the constant out:

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

= 13 \cdot \frac{1}{2} \cdot \int cos (5 π x) + cos (- 3 π x) d x

Simplify

cos (5 π x) + cos (- 3 π x) : cos (3 π x) + cos (5 π x)

= 13 \cdot \frac{1}{2} \cdot \int cos (3 π x) + cos (5 π 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

= 13 \cdot \frac{1}{2} (\int cos (3 π x) d x + \int cos (5 π x) d x)

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

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

= 13 \cdot \frac{1}{2} (\frac{1}{3 π} sin (3 π x) + \frac{1}{5 π} sin (5 π x))

Simplify

13 \cdot \frac{1}{2} (\frac{1}{3 π} sin (3 π x) + \frac{1}{5 π} sin (5 π x)) : \frac{13}{2} (\frac{1}{3 π} sin (3 π x) + \frac{1}{5 π} sin (5 π x))

= \frac{13}{2} (\frac{1}{3 π} sin (3 π x) + \frac{1}{5 π} sin (5 π x))

Add a constant to the solution

= \frac{13}{2} (\frac{1}{3 π} sin (3 π x) + \frac{1}{5 π} sin (5 π x)) + C

n = 0 \sum \infty cos (7 πn)

\int t (t^{2} - 3 t + 5) d t

f (x) = \frac{3}{x + 1}, at x = 2

\int sin^{2} (θ) cos^{5} (θ) d θ

\frac{\partial}{\partial x} (e^{5 x y})

What is the integral of 13cos(pix)cos(4pix) ?
The integral of 13cos(pix)cos(4pix) is 13/2 (1/(3pi)sin(3pix)+1/(5pi)sin(5pix))+C