To solve the integral:

$\int_{0}^{\frac{\pi}{2}} \cos^5(x) \, dx$

we'll go through a step-by-step solution and explanation.

Step 1: Rewrite the Integrand

The integrand, $\cos^5(x)$, is an odd power of cosine, which we can rewrite as: $\cos^5(x) = \cos^4(x) \cdot \cos(x) = (\cos^2(x))^2 \cdot \cos(x)$

Using the Pythagorean identity $\cos^2(x) = 1 - \sin^2(x)$, we get: $\cos^4(x) = (1 - \sin^2(x))^2$

Thus, $\cos^5(x) = (1 - \sin^2(x))^2 \cos(x)$

Step 2: Substitute $u = \sin(x)$

Let $u = \sin(x)$. Then, $du = \cos(x) \, dx$.

When $x = 0$, $u = \sin(0) = 0$. When $x = \frac{\pi}{2}$, $u = \sin\left(\frac{\pi}{2}\right) = 1$.

Therefore, the integral becomes: $\int_{0}^{\frac{\pi}{2}} \cos^5(x) \, dx = \int_{0}^{1} (1 - u^2)^2 \, du$

Step 3: Expand $(1 - u^2)^2$

Expanding $(1 - u^2)^2$, we get: $(1 - u^2)^2 = 1 - 2u^2 + u^4$

So the integral now becomes: $\int_{0}^{1} (1 - 2u^2 + u^4) \, du$

Step 4: Integrate Term by Term

Now we integrate each term with respect to $u$: $\int_{0}^{1} (1 - 2u^2 + u^4) \, du = \int_{0}^{1} 1 \, du - 2 \int_{0}^{1} u^2 \, du + \int_{0}^{1} u^4 \, du$

1. First integral: $\int_{0}^{1} 1 \, du = u \Big|_{0}^{1} = 1$.

2. Second integral: $-2 \int_{0}^{1} u^2 \, du = -2 \cdot \frac{u^3}{3} \Big|_{0}^{1} = -2 \cdot \frac{1}{3} = -\frac{2}{3}$.

3. Third integral: $\int_{0}^{1} u^4 \, du = \frac{u^5}{5} \Big|_{0}^{1} = \frac{1}{5}$.

Step 5: Combine the Results

Now, we add the results from each part: $\int_{0}^{1} (1 - 2u^2 + u^4) \, du = 1 - \frac{2}{3} + \frac{1}{5}$

To simplify, we find a common denominator of 15: $1 = \frac{15}{15}, \quad -\frac{2}{3} = -\frac{10}{15}, \quad \frac{1}{5} = \frac{3}{15}$

So, $1 - \frac{2}{3} + \frac{1}{5} = \frac{15 - 10 + 3}{15} = \frac{8}{15}$

Final Answer

$\int_{0}^{\frac{\pi}{2}} \cos^5(x) \, dx = \frac{8}{15}$

Would you like further details on any part of this solution?

Questions to Explore Further:

1. How would the solution change if we had $\cos^6(x)$ instead?
2. Why do we use trigonometric identities in integration?
3. What is the importance of substitution in integrals like this?
4. Can we solve this integral using numerical methods?
5. How do definite integrals differ in their setup compared to indefinite integrals?

Tip:

For integrals involving odd powers of trigonometric functions, breaking down the powers and