Let's evaluate the given indefinite integrals:

(a) $\int \cos^5(x) \, dx$

To solve $\int \cos^5(x) \, dx$, we can use the identity $\cos^2(x) = 1 - \sin^2(x)$ and a substitution method. Here’s the step-by-step solution:

1. Rewrite $\cos^5(x)$ as $\cos^4(x) \cdot \cos(x)$. $\int \cos^5(x) \, dx = \int \cos^4(x) \cdot \cos(x) \, dx$
2. Express $\cos^4(x)$ as $(\cos^2(x))^2$ and then apply the identity $\cos^2(x) = 1 - \sin^2(x)$: $= \int (1 - \sin^2(x))^2 \cdot \cos(x) \, dx$
3. Let $u = \sin(x)$, then $du = \cos(x) \, dx$. Substitute $u$ and $du$ into the integral: $= \int (1 - u^2)^2 \, du$
4. Expand $(1 - u^2)^2$: $= \int (1 - 2u^2 + u^4) \, du$
5. Integrate each term separately: $= \int 1 \, du - 2 \int u^2 \, du + \int u^4 \, du$
6. Calculate each integral: $= u - \frac{2u^3}{3} + \frac{u^5}{5} + C$
7. Substitute back $u = \sin(x)$: $= \sin(x) - \frac{2 \sin^3(x)}{3} + \frac{\sin^5(x)}{5} + C$

So, the answer for part (a) is: $\int \cos^5(x) \, dx = \sin(x) - \frac{2 \sin^3(x)}{3} + \frac{\sin^5(x)}{5} + C$

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(b) $\int 4 \sin^4(x) \, dx$

To evaluate $\int 4 \sin^4(x) \, dx$, we can use the power-reduction formula.

1. Use the power-reduction identity for $\sin^2(x)$: $\sin^2(x) = \frac{1 - \cos(2x)}{2}$

2. Therefore, $\sin^4(x) = \left(\frac{1 - \cos(2x)}{2}\right)^2$. $\sin^4(x) = \frac{1}{4}(1 - 2\cos(2x) + \cos^2(2x))$

3. Substitute this into the integral: $\int 4 \sin^4(x) \, dx = \int 4 \cdot \frac{1}{4}(1 - 2\cos(2x) + \cos^2(2x)) \, dx$ Simplify by distributing $4$: $= \int (1 - 2\cos(2x) + \cos^2(2x)) \, dx$

4. Now, split the integral: $= \int 1 \, dx - 2 \int \cos(2x) \, dx + \int \cos^2(2x) \, dx$

5. Integrate each term:

   • The first term is straightforward: $\int 1 \, dx = x$
   • For the second term, use $\int \cos(2x) \, dx = \frac{\sin(2x)}{2}$: $-2 \int \cos(2x) \, dx = -\sin(2x)$
   • For the third term, use the power-reduction formula again for $\cos^2(2x)$: $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$ Thus, $\int \cos^2(2x) \, dx = \int \frac{1 + \cos(4x)}{2} \, dx = \frac{x}{2} + \frac{\sin(4x)}{8}$

6. Combine all the terms: $= x - \sin(2x) + \frac{x}{2} + \frac{\sin(4x)}{8} + C$ Simplify: $= \frac{3x}{2} - \sin(2x) + \frac{\sin(4x)}{8} + C$

So, the answer for part (b) is: $\int 4 \sin^4(x) \, dx = \frac{3x}{2} - \sin(2x) + \frac{\sin(4x)}{8} + C$

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Would you like a further breakdown of any steps?

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Related Questions:

1. How can trigonometric identities help in simplifying integrals?
2. What are the power-reduction formulas and when are they used in integration?
3. How does the substitution method simplify complex trigonometric integrals?
4. What are common applications of trigonometric integrals in physics?
5. How can you determine if a trigonometric integral requires substitution or another technique?

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Tip: When dealing with powers of sine and cosine in integrals, look for ways to express even powers in terms of half-angles to simplify the process.