What is the integral of 9t^2sin(t^3) ?

The integral of 9t^2sin(t^3) is -3cos(t^3)+C

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integral of 9t^2sin(t^3)

\int 9 t^{2} sin (t^{3}) d t

- 3 cos (t^{3}) + C

Solution steps

\int 9 t^{2} sin (t^{3}) d t

Take the constant out:

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

= 9 \cdot \int t^{2} sin (t^{3}) d t

Apply u-substitution

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

= 9 \cdot \int \frac{sin ( u )}{3} d u

Take the constant out:

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

= 9 \cdot \frac{1}{3} \cdot \int sin (u) d u

Use the common integral:

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

= 9 \cdot \frac{1}{3} (- cos (u))

Substitute back

u = t^{3}

= 9 \cdot \frac{1}{3} (- cos (t^{3}))

Simplify

9 \cdot \frac{1}{3} (- cos (t^{3})) : - 3 cos (t^{3})

= - 3 cos (t^{3})

Add a constant to the solution

= - 3 cos (t^{3}) + C

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What is the integral of 9t^2sin(t^3) ?
The integral of 9t^2sin(t^3) is -3cos(t^3)+C