What is the integral of sin(8pit) ?

The integral of sin(8pit) is -1/(8pi)cos(8pit)+C

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integral of sin(8pit)

\int sin (8 π t) d t

- \frac{1}{8 π} cos (8 π t) + C

Solution steps

\int sin (8 π t) d t

Apply u-substitution

: \int sin (u) \frac{1}{8 π} d u

= \int sin (u) \frac{1}{8 π} d u

Take the constant out:

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

= \frac{1}{8 π} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{8 π} (- cos (u))

Substitute back

u = 8 π t

= \frac{1}{8 π} (- cos (8 π t))

Simplify

= - \frac{1}{8 π} cos (8 π t)

Add a constant to the solution

= - \frac{1}{8 π} cos (8 π t) + C

\frac{d}{d x} (x ln (2 x + 3))

x y^{^{'}} = y + 2 x y

(1 + x)^{2}

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

What is the integral of sin(8pit) ?
The integral of sin(8pit) is -1/(8pi)cos(8pit)+C