The given integral is:

$I = \int x^5 \cos(x^3) \, dx$

To solve this, we can use a substitution method.

Step 1: Substitution

Let $u = x^3$, which simplifies the powers of $x$ inside the cosine function.

• Then, $du = 3x^2 \, dx$, or equivalently, $\frac{du}{3} = x^2 \, dx$.

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

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

Thus, the integral becomes:

$I = \int x^5 \cos(x^3) \, dx = \int u \cdot x^2 \cos(u) \cdot \frac{du}{3}$

Step 2: Simplification

This simplifies to:

$I = \frac{1}{3} \int u \cos(u) \, du$

Step 3: Integration by Parts

We now solve the integral $\int u \cos(u) \, du$ using integration by parts, where we let:

• $v = u$, so $dv = du$,
• $dw = \cos(u) \, du$, so $w = \sin(u)$.

Using the integration by parts formula:

$\int v \, dw = v \cdot w - \int w \, dv$

We have:

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

The integral of $\sin(u)$ is $-\cos(u)$, so:

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

Step 4: Substitution Back

Thus, the original integral becomes:

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

Substitute back $u = x^3$:

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

Final Answer:

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

Let me know if you'd like further clarification or details!

Related Questions:

1. How does integration by parts work for more complex functions?
2. What other integrals involving trigonometric functions can be solved using substitution?
3. Can substitution always simplify integrals involving composite functions?
4. What are the steps for solving integrals that involve products of functions?
5. How do definite integrals differ in their solutions compared to indefinite integrals?

Tip: Always look for opportunities to apply substitution in integrals where you have composite functions like $\cos(x^3)$.