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

The integral of-1/(n^2)cos(nt) is -1/(n^3)sin(nt)+C

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integral of-1/(n^2)cos(nt)

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

- \frac{1}{n ^{3}} sin (n t) + C

Solution steps

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

Apply u-substitution

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

= - \frac{1}{n ^{2}} \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 ^{2}} \cdot \frac{1}{n} \cdot \int cos (u) d u

Use the common integral:

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

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

Substitute back

u = n t

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

Simplify

- \frac{1}{n ^{2}} \cdot \frac{1}{n} sin (n t) : - \frac{1}{n ^{3}} sin (n t)

= - \frac{1}{n ^{3}} sin (n t)

Add a constant to the solution

= - \frac{1}{n ^{3}} sin (n t) + C

\frac{\partial}{\partial x} (x^{3} cos (x y))

x = y^{2}, x = 6 - 2 y^{2}

\int (9 x 1 + 2 x^{2}) d x

x \to 0 lim (\frac{x ^{2} - x}{x})

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

What is the integral of-1/(n^2)cos(nt) ?
The integral of-1/(n^2)cos(nt) is -1/(n^3)sin(nt)+C