We are tasked with evaluating the integral:

$\int \cos(3x + 2) \sin^4(3x + 2) \, dx$

Step 1: Substitution

Let us start by making the substitution: $u = 3x + 2$ This gives: $du = 3 \, dx \quad \Rightarrow \quad dx = \frac{du}{3}$

Substituting into the integral: $\int \cos(u) \sin^4(u) \frac{du}{3}$

Factoring out the constant: $\frac{1}{3} \int \cos(u) \sin^4(u) \, du$

Step 2: Simplifying with Power Reduction

We can express $\sin^4(u)$ using the identity $\sin^2(u) = 1 - \cos^2(u)$: $\sin^4(u) = (\sin^2(u))^2 = (1 - \cos^2(u))^2$

Expanding $(1 - \cos^2(u))^2$: $\sin^4(u) = 1 - 2\cos^2(u) + \cos^4(u)$

Thus, the integral becomes: $\frac{1}{3} \int \cos(u) \left(1 - 2\cos^2(u) + \cos^4(u)\right) \, du$

Step 3: Breaking the Integral into Three Terms

We can now split the integral into three simpler integrals: $\frac{1}{3} \left( \int \cos(u) \, du - 2 \int \cos^3(u) \, du + \int \cos^5(u) \, du \right)$

Integral 1: $\int \cos(u) \, du$

This is straightforward: $\int \cos(u) \, du = \sin(u)$

Integral 2: $\int \cos^3(u) \, du$

Using the identity $\cos^3(u) = \cos(u)(1 - \sin^2(u))$, we get: $\int \cos^3(u) \, du = \int \cos(u) \, du - \int \cos(u) \sin^2(u) \, du$ The first part is $\sin(u)$, while the second part requires a substitution. However, this integral follows a known result: $\int \cos^3(u) \, du = \frac{\sin(u)}{3} + \frac{\sin^3(u)}{3}$

Integral 3: $\int \cos^5(u) \, du$

Similarly, we use the identity $\cos^5(u) = \cos(u)(1 - \sin^2(u))^2$, and the result for this is: $\int \cos^5(u) \, du = \frac{\sin(u)}{5} + \frac{\sin^3(u)}{5} + \frac{\sin^5(u)}{5}$

Step 4: Combine Results

Now, combining all the terms, we have:

$\frac{1}{3} \left( \sin(u) - 2 \left( \frac{\sin(u)}{3} + \frac{\sin^3(u)}{3} \right) + \left( \frac{\sin(u)}{5} + \frac{\sin^3(u)}{5} + \frac{\sin^5(u)}{5} \right) \right)$

Finally, substitute $u = 3x + 2$ back into the result.

Would you like me to simplify further or provide details on any of these steps?

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Related questions:

1. How do you solve integrals involving trigonometric identities?
2. What are common techniques for solving trigonometric integrals?
3. Can substitution always be applied for integrals like this?
4. How is the power reduction formula used in integration?
5. What are the differences between definite and indefinite integrals involving trigonometric functions?

Tip: Use trigonometric identities to simplify complex integrals involving powers of sine and cosine.