What is the integral of sin(3xy) ?

The integral of sin(3xy) is -1/(3y)cos(3yx)+C

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integral of sin(3xy)

\int sin (3 x y) d x

- \frac{1}{3 y} cos (3 y x) + C

Solution steps

\int sin (3 x y) d x

Apply u-substitution

: \int \frac{sin ( u )}{3 y} d u

= \int \frac{sin ( u )}{3 y} d u

Take the constant out:

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

= \frac{1}{3 y} \cdot \int sin (u) d u

Use the common integral:

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

= \frac{1}{3 y} (- cos (u))

Substitute back

u = 3 y x

= \frac{1}{3 y} (- cos (3 y x))

Simplify

= - \frac{1}{3 y} cos (3 y x)

Add a constant to the solution

= - \frac{1}{3 y} cos (3 y x) + C

f (x) = x + x

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

y = - x^{2} - x, y = 3 x

(\frac{d x}{d y}) = k (a - x) (a - x)

\frac{1}{12 x}

What is the integral of sin(3xy) ?
The integral of sin(3xy) is -1/(3y)cos(3yx)+C