What is the integral of (cos(3pix))/2 ?

The integral of (cos(3pix))/2 is 1/(6pi)sin(3pix)+C

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integral of (cos(3pix))/2

\int \frac{cos ( 3 π x )}{2} d x

\frac{1}{6 π} sin (3 π x) + C

Solution steps

\int \frac{cos ( 3 π 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 cos (3 π x) d x

Apply u-substitution

: \int cos (u) \frac{1}{3 π} d u

= \frac{1}{2} \cdot \int cos (u) \frac{1}{3 π} d u

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{3 π} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{2} \cdot \frac{1}{3 π} sin (u)

Substitute back

u = 3 π x

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

Simplify

\frac{1}{2} \cdot \frac{1}{3 π} sin (3 π x) : \frac{1}{6 π} sin (3 π x)

= \frac{1}{6 π} sin (3 π x)

Add a constant to the solution

= \frac{1}{6 π} sin (3 π x) + C

\frac{\partial}{\partial x} ((3 x - 2 y)^{2})

\frac{d}{d n} (\frac{1}{n})

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

6 x e^{5 x}

\frac{d}{d x} (\frac{3 ^{x}}{3 ^{x} + 2})

What is the integral of (cos(3pix))/2 ?
The integral of (cos(3pix))/2 is 1/(6pi)sin(3pix)+C