What is the integral of-1/n cos(nx) ?

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

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integral of-1/n cos(nx)

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

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

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

= - \frac{1}{n} \cdot \int \frac{cos ( u )}{n} d u

Take the constant out:

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

= - \frac{1}{n} \cdot \frac{1}{n} \cdot \int cos (u) d u

Use the common integral:

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

= - \frac{1}{n} \cdot \frac{1}{n} sin (u)

Substitute back

u = n x

= - \frac{1}{n} \cdot \frac{1}{n} sin (n x)

Simplify

- \frac{1}{n} \cdot \frac{1}{n} sin (n x) : - \frac{1}{n ^{2}} sin (n x)

= - \frac{1}{n ^{2}} sin (n x)

Add a constant to the solution

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

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