All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
$cos ^6 x$

$= (cos x)^6$

$\approx (1 -\frac{x^2}{2} + \frac{x^4}{24})^6$

$= 1 + {6 \choose 1}(1)(-\frac{x^2}{2} + \frac{x^4}{24}) + {6 \choose 2}(1)(-\frac{x^2}{2} + \frac{x^4}{24})^2 + \ldots$

$= 1 + 6(-\frac{x^2}{2} + \frac{x^4}{24}) + 15(-\frac{x^2}{2} + \frac{x^4}{24})^2 + \ldots$

$\approx 1 - 3x^2 + 4x^4$

(ii)
(a)

$\int_0^a g(x) dx$

$= \int_0^a 1 - 3x^2 + 4x^4 dx$

$= x - x^3 + \frac{4x^5}{5} \biggl|_0^a$

$= a - a^3 + \frac{4a^5}{5}$

When $a = \frac{\pi}{4}, ~ \int_0^a g(x) dx \approx 0.540$

(b)
Using the Graphing Calculator, $\int_0^{\frac{\pi}{4}} g(x) dx = 0.475$

The approximation is not very good since $\frac{\pi}{4}$ is not very close to zero.
We only considered up to $x^4$, thus the approximation can be improved by taking into consideration more terms.

### KS Comments:

Just don’t be careless for (i)! Some students still can be unsure of how to evaluate a definite integral using Graphing Calculator. Lastly, maclaurin’s series expansion are most accurate when value is close to one since they are centred at x = 0.

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