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

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

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integral of x^2sin^2(x^3)

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

\frac{1}{6} (x^{3} - \frac{1}{2} sin (2 x^{3})) + C

Solution steps

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

Apply u-substitution

= \int \frac{sin ^{2} ( u )}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sin^{2} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \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}{3} \cdot \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}{3} \cdot \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}{3} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u))

Substitute back

u = x^{3}

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

Simplify

\frac{1}{3} \cdot \frac{1}{2} (x^{3} - \frac{1}{2} sin (2 x^{3})) : \frac{1}{6} (x^{3} - \frac{1}{2} sin (2 x^{3}))

= \frac{1}{6} (x^{3} - \frac{1}{2} sin (2 x^{3}))

Add a constant to the solution

= \frac{1}{6} (x^{3} - \frac{1}{2} sin (2 x^{3})) + C

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