[M] (Calculus required) Recall from calculus that integrals such as

\(\int {\left( {{\bf{5co}}{{\bf{s}}^{\bf{3}}}t - {\bf{6co}}{{\bf{s}}^{\bf{4}}}t + {\bf{5co}}{{\bf{s}}^{\bf{5}}}t - {\bf{12co}}{{\bf{s}}^{\bf{6}}}t} \right){\bf{d}}t} \) (10)

Are tedious to compute. (The usual method is to apply integration by parts repeteatedly and use the half angle formula.) Use the matrix P or \({P^{ - {\bf{1}}}}\) from Exercise 17 to transform (10); then compute the integral.

The C coordinate of the integrand is \(\left( {0,0,0,5, - 6,5, - 12} \right)\).

By the inverse matrix \({P^{ - 1}} = \frac{1}{{32}}\left[ {\begin{array}{*{20}{c}}{32}&0&{16}&0&{12}&0&{10}\\0&{32}&0&{24}&0&{20}&0\\0&0&{16}&0&{16}&0&{15}\\0&0&0&8&0&{10}&0\\0&0&0&0&4&0&6\\0&0&0&0&0&2&0\\0&0&0&0&0&0&1\end{array}} \right]\), the B coordinate is found as

\(\begin{aligned} {P^{ - 1}}\left( {0,\,0,\,0,\,5,\, - 6,\,5,\, - 12} \right) &= \frac{1}{{32}}\left[ {\begin{array}{*{20}{c}}{32}&0&{16}&0&{12}&0&{10}\\0&{32}&0&{24}&0&{20}&0\\0&0&{16}&0&{16}&0&{15}\\0&0&0&8&0&{10}&0\\0&0&0&0&4&0&6\\0&0&0&0&0&2&0\\0&0&0&0&0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{c}}0\\0\\0\\5\\{ - 6}\\5\\{ - 12}\end{array}} \right]\\ &= \left[ {\begin{array}{*{20}{c}}{ - 6}&{\frac{{55}}{8}}&{ - \frac{{69}}{8}}&{\frac{{45}}{{16}}}&{ - 3}&{\frac{5}{{16}}}&{ - \frac{3}{8}}\end{array}} \right].\end{aligned}\)

Matrix P is

\(P = \left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&0&1&0&{ - 1}\\0&1&0&{ - 3}&0&5&0\\0&0&2&0&{ - 8}&0&{18}\\0&0&0&4&0&{ - 20}&0\\0&0&0&0&8&0&{ - 48}\\0&0&0&0&0&{16}&0\\0&0&0&0&0&0&{32}\end{array}} \right]\).

Using the B coordinate, the given integral form can be written as

\(\int {\left( { - 6 + \frac{{55}}{8}\cos t - \frac{{69}}{8}\cos 2t + \frac{{45}}{{16}}\cos 3t - 3\cos 4t + \frac{5}{{16}}\cos 5t - \frac{3}{8}\cos 6t} \right){\rm{d}}t} \).

Use the following code forintegration:

\(\begin{array}{c} > > {\rm{fun}} = - 6 + \frac{{55}}{8}\cos t - \frac{{69}}{8}\cos 2t + \frac{{45}}{{16}}\cos 3t - 3\cos 4t + \frac{5}{{16}}\cos 5t - \frac{3}{8}\cos 6t;\\ > > F = {\mathop{\rm int}} ({\rm{fun}});\end{array}\)

So, the value of integral is

\( - 6t + \frac{{55}}{8}\sin t - \frac{{69}}{{16}}\sin \left( {2t} \right) + \frac{{15}}{{16}}\sin \left( {3t} \right) - \frac{3}{4}\sin \left( {4t} \right) + \frac{1}{{16}}\sin \left( {5t} \right) - \frac{1}{{16}}\sin \left( {6t} \right) + C\).