What is the integral of (x^2-6x)sin^2((x^3)/3-3x^2) ?

The integral of (x^2-6x)sin^2((x^3)/3-3x^2) is 1/2 ((x^3)/3-3x^2-1/2 sin(2((x^3)/3-3x^2)))+C

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integral of (x^2-6x)sin^2((x^3)/3-3x^2)

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

\frac{1}{2} (\frac{x ^{3}}{3} - 3 x^{2} - \frac{1}{2} sin (2 (\frac{x ^{3}}{3} - 3 x^{2}))) + C

Solution steps

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

Apply u-substitution

= \int sin^{2} (u) d u

Rewrite using trig identities

= \int \frac{1 - cos ( 2 u )}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 - cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int 1 d u - \int cos (2 u) d u)

\int 1 d u = u

\int cos (2 u) d u = \frac{1}{2} sin (2 u)

= \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = \frac{x ^{3}}{3} - 3 x^{2}

= \frac{1}{2} (\frac{x ^{3}}{3} - 3 x^{2} - \frac{1}{2} sin (2 (\frac{x ^{3}}{3} - 3 x^{2})))

Add a constant to the solution

= \frac{1}{2} (\frac{x ^{3}}{3} - 3 x^{2} - \frac{1}{2} sin (2 (\frac{x ^{3}}{3} - 3 x^{2}))) + C

What is the integral of (x^2-6x)sin^2((x^3)/3-3x^2) ?
The integral of (x^2-6x)sin^2((x^3)/3-3x^2) is 1/2 ((x^3)/3-3x^2-1/2 sin(2((x^3)/3-3x^2)))+C