What is the integral of 1/n sin(nt) ?

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

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integral of 1/n sin(nt)

\int \frac{1}{n} sin (n t) d t

- \frac{1}{n ^{2}} cos (n t) + C

Solution steps

\int \frac{1}{n} sin (n t) d t

Take the constant out:

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

= \frac{1}{n} \cdot \int sin (n t) d t

Apply u-substitution

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

Use the common integral:

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

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

Substitute back

u = n t

= \frac{1}{n} \cdot \frac{1}{n} (- cos (n t))

Simplify

\frac{1}{n} \cdot \frac{1}{n} (- cos (n t)) : - \frac{1}{n ^{2}} cos (n t)

= - \frac{1}{n ^{2}} cos (n t)

Add a constant to the solution

= - \frac{1}{n ^{2}} cos (n t) + C

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