What is the integral of sin(9pit) ?

The integral of sin(9pit) is -1/(9pi)cos(9pit)+C

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integral of sin(9pit)

\int sin (9 π t) d t

- \frac{1}{9 π} cos (9 π t) + C

Solution steps

\int sin (9 π t) d t

Apply u-substitution

: \int sin (u) \frac{1}{9 π} d u

= \int sin (u) \frac{1}{9 π} d u

Take the constant out:

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

= \frac{1}{9 π} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{9 π} (- cos (u))

Substitute back

u = 9 π t

= \frac{1}{9 π} (- cos (9 π t))

Simplify

= - \frac{1}{9 π} cos (9 π t)

Add a constant to the solution

= - \frac{1}{9 π} cos (9 π t) + C

x \to 3 lim (\frac{x - 5}{x - 3})

\frac{d}{d x} (\frac{ln ( x )}{x ^{5}})

e^{l n (x)}

\frac{2 x}{( x ^{3} - 6 ) ^{2}}

\frac{d}{d x} (\frac{sin ^{2} ( x )}{x})

What is the integral of sin(9pit) ?
The integral of sin(9pit) is -1/(9pi)cos(9pit)+C