To solve this indefinite integral by substitution, let's proceed step-by-step:

Problem

$\int \cos^7(8x) \sin(8x) \, dx$

Step 1: Choose a Substitution

Let's set: $u = \cos(8x)$ Then, the derivative of $u$ with respect to $x$ is: $\frac{du}{dx} = -8 \sin(8x)$ or rearranging, $\sin(8x) \, dx = -\frac{1}{8} \, du$

Step 2: Substitute in Terms of $u$

Rewrite the integral using $u$: $\int \cos^7(8x) \sin(8x) \, dx = \int u^7 \cdot \left(-\frac{1}{8}\right) \, du$ This simplifies to: $-\frac{1}{8} \int u^7 \, du$

Step 3: Integrate with Respect to $u$

Now integrate $u^7$ with respect to $u$: $-\frac{1}{8} \cdot \frac{u^8}{8} = -\frac{1}{64} u^8 + C$

Step 4: Substitute Back in Terms of $x$

Since $u = \cos(8x)$, substitute back: $-\frac{1}{64} \cos^8(8x) + C$

Final Answer

$\int \cos^7(8x) \sin(8x) \, dx = -\frac{1}{64} \cos^8(8x) + C$

Would you like further clarification on any of the steps?

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Here are some related questions to consider:

1. How would the solution change if the power of cosine were different?
2. What if the integral involved $\cos^7(8x) \cos(8x)$ instead?
3. Can you solve similar integrals using trigonometric identities instead of substitution?
4. How does the constant $C$ affect the family of solutions?
5. What techniques can be used for integrals without obvious substitutions?

Tip: Always double-check substitutions by differentiating back to verify if the solution matches the original integrand.