Let's go through each part of the question one by one.

Part (a)

To obtain the Maclaurin series for $\cos^4 x$, you can start with the Maclaurin series for $\cos x$, which is: $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \dots$ Then, $\cos^4 x$ can be expanded by raising this series to the fourth power and simplifying terms as needed. This involves binomial expansion or re-expanding the product, but since we don’t need to actually work it out here, it suffices to state that $\cos^4 x$ can be approximated by expanding the series for $\cos x$ and then taking the fourth power of the truncated series.

The interval of convergence for the Maclaurin series of $\cos x$ is $(-\infty, \infty)$, meaning it converges for all real $x$. This interval would apply to $\cos^4 x$ as well, assuming we use a truncated version of the series for approximation.

Part (b)

To approximate the integral $\int_0^1 \cos^4 x \, dx$, we substitute $\cos^4 x$ with the terms of its Maclaurin series from part (a). For example, using the first few terms might give us an approximation of $\cos^4 x$ as: $\cos^4 x \approx 1 - 2x^2 + \frac{5}{3} x^4$ Then, we can integrate this polynomial term by term over the interval $[0, 1]$: $\int_0^1 \left(1 - 2x^2 + \frac{5}{3} x^4\right) \, dx$ Calculating each part: $\int_0^1 1 \, dx = 1$ $\int_0^1 2x^2 \, dx = \frac{2}{3}$ $\int_0^1 \frac{5}{3} x^4 \, dx = \frac{5}{15} = \frac{1}{3}$ Thus, the approximation becomes: $1 - \frac{2}{3} + \frac{1}{3} = \frac{2}{3} \approx 0.667$

Part (c)

The exact value of the integral is given as $0.579$. The approximation obtained in part (b) is not very close to this value. This discrepancy occurs because the series approximation for $\cos^4 x$ may not have included enough terms to accurately represent $\cos^4 x$ over the interval $[0, 1]$.

To improve accuracy, we could include more terms in the Maclaurin series for $\cos^4 x$ to provide a better approximation, or use a different approach (such as numerical integration methods) to approximate the integral more accurately.

Part (d)

For the integral $\int_5^6 \cos^4 x \, dx$, using a Maclaurin series may not be appropriate. The Maclaurin series is an approximation around $x = 0$ and converges well near zero. However, since the interval $[5, 6]$ is far from zero, the Maclaurin series may not approximate $\cos^4 x$ well in this range. A Taylor series centered closer to the interval $[5, 6]$ or a numerical integration method would be more appropriate in this case.

Would you like more details on any specific part or further help on this problem?

Here are some related questions for further exploration:

1. How would you derive the Maclaurin series for $\sin^2 x$ or $\cos^2 x$?
2. What other methods could be used to approximate integrals beyond Taylor and Maclaurin series?
3. Can a Taylor series centered at a different point be used for integration over different intervals?
4. How does the error in approximation depend on the number of terms used in the series?
5. What are some common applications of Taylor and Maclaurin series in physics or engineering?

Tip: When working with Maclaurin or Taylor series, always consider the interval over which you want to approximate, as series centered at $x = 0$ may not perform well far from zero.