What is the integral of sin(9pix) ?

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

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integral of sin(9pix)

\int sin (9 π x) d x

- \frac{1}{9 π} cos (9 π x) + C

Solution steps

\int sin (9 π x) d x

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 x

= \frac{1}{9 π} (- cos (9 π x))

Simplify

= - \frac{1}{9 π} cos (9 π x)

Add a constant to the solution

= - \frac{1}{9 π} cos (9 π x) + C

\int (x^{3} e^{x^{2}}) d x

y^{^{''}} + 16 y = 4 csc^{2} (4 t)

\frac{d}{d x} ((x^{4} + 1)^{\frac{1}{4}})

\frac{d}{d x} (\frac{3 x ^{4} + 7}{x ^{2}})

x \to 0 lim (8)

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