What is the integral of 18θsin(18θ^2) ?

The integral of 18θsin(18θ^2) is -1/2 cos(18θ^2)+C

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integral of 18θsin(18θ^2)

\int 18 θ sin (18 θ^{2}) d θ

- \frac{1}{2} cos (18 θ^{2}) + C

Solution steps

\int 18 θ sin (18 θ^{2}) d θ

Take the constant out:

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

= 18 \cdot \int θ sin (18 θ^{2}) d θ

Apply u-substitution

: \int \frac{sin ( u )}{36} d u

= 18 \cdot \int \frac{sin ( u )}{36} d u

Take the constant out:

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

= 18 \cdot \frac{1}{36} \cdot \int sin (u) d u

Use the common integral:

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

= 18 \cdot \frac{1}{36} (- cos (u))

Substitute back

u = 18 θ^{2}

= 18 \cdot \frac{1}{36} (- cos (18 θ^{2}))

Simplify

18 \cdot \frac{1}{36} (- cos (18 θ^{2})) : - \frac{1}{2} cos (18 θ^{2})

= - \frac{1}{2} cos (18 θ^{2})

Add a constant to the solution

= - \frac{1}{2} cos (18 θ^{2}) + C

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What is the integral of 18θsin(18θ^2) ?
The integral of 18θsin(18θ^2) is -1/2 cos(18θ^2)+C