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

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

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integral of 1/n sin(nx)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{sin ( u )}{n} d u

= \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 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{104}{s ^{2} + 8 s + 52}

\frac{d}{d x} (\frac{x ^{3} - 7}{x ^{2} + 2})

\frac{d ^{3}}{d x ^{3}} (\frac{1}{1 - x})

\frac{d}{d x} (3 x^{2} + 2 x - 4)

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