Find the integral $$ \int x^{5} \cos \left(x^{3}\right) d x=\square $$

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To evaluate the integral:

$$
\int x^{5} \cos \left(x^{3}\right) \, dx
$$

we can use substitution and integration by parts.

**Step 1: Substitution**

Let $$ u = x^3 $$. Then, the derivative $$ du $$ is:

$$
du = 3x^2 \, dx \quad \Rightarrow \quad x^2 \, dx = \frac{du}{3}
$$

Now, express $$ x^5 $$ in terms of $$ u $$:

$$
x^5 = x^3 \cdot x^2 = u \cdot x^2
$$

Substituting into the integral:

$$
\int x^{5} \cos \left(x^{3}\right) \, dx = \int u \cos(u) \cdot \frac{du}{3} = \frac{1}{3} \int u \cos(u) \, du
$$

**Step 2: Integration by Parts**

To evaluate $$ \int u \cos(u) \, du $$, we use integration by parts:

$$
\text{Let} \quad
\begin{cases}
v = u \
dw = \cos(u) \, du
\end{cases}
$$

Then,

$$
dv = du \quad \text{and} \quad w = \sin(u)
$$

Applying the integration by parts formula $$ \int v \, dw = v w - \int w \, dv $$:

$$
\int u \cos(u) \, du = u \sin(u) - \int \sin(u) \, du = u \sin(u) + \cos(u) + C
$$

**Step 3: Substitute Back**

Substitute $$ u = x^3 $$ back into the expression:

$$
\frac{1}{3} \left( u \sin(u) + \cos(u) \right) + C = \frac{1}{3} \left( x^3 \sin(x^3) + \cos(x^3) \right) + C
$$

**Final Answer:**

$$
\int x^{5} \cos \left(x^{3}\right) \, dx = \frac{x^{3} \sin(x^{3}) + \cos(x^{3})}{3} + C
$$

where $$ C $$ is the constant of integration.
The integral of $$ x^{5} \cos \left(x^{3}\right) $$ with respect to $$ x $$ is: $$ \frac{x^{3} \sin(x^{3}) + \cos(x^{3})}{3} + C $$ where $$ C $$ is the constant of integration.

Find the integral

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