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

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

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integral of-1/n cos(nt)

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

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

Solution steps

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

Apply u-substitution

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

= - \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 t

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

Simplify

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

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

Add a constant to the solution

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

x^{2} + 3 x

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

\int x^{4} - 2 x^{3} + x^{2} d x

f (x) = 4 e^{2 x}

F (x) = (3 x^{6} + 4 x^{3})^{4}

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