What is the integral of sin(npit) ?

The integral of sin(npit) is -1/(pin)cos(pint)+C

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integral of sin(npit)

\int sin (nπ t) d t

- \frac{1}{πn} cos (πn t) + C

Solution steps

\int sin (nπ t) d t

Apply u-substitution

: \int \frac{sin ( u )}{πn} d u

= \int \frac{sin ( u )}{πn} d u

Take the constant out:

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

= \frac{1}{πn} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{πn} (- cos (u))

Substitute back

u = nπ t

= \frac{1}{πn} (- cos (nπ t))

Simplify

= - \frac{1}{πn} cos (πn t)

Add a constant to the solution

= - \frac{1}{πn} cos (πn t) + C

x \cdot \frac{d y}{d x} + 6 y = x^{3} - x

\frac{d}{d t} (4)

(\frac{2 t}{t ^{2} + 1})^{^{'}}

\frac{\partial}{\partial y} (x^{3} - 5 x y)

x \to 0 + lim (sin (x) ln (2 x))

What is the integral of sin(npit) ?
The integral of sin(npit) is -1/(pin)cos(pint)+C