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[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.

Short Answer

Expert verified

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

Step by step solution

01

Write the coordinate of the integrand

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

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

02

Change the integral using the B coordinate

Matrix P is

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

03

Find the inverse of P

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

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

04

Integrate the integral using MATLAB

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

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

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Most popular questions from this chapter

Let \(B = \left\{ {\left( {\begin{array}{*{20}{c}}{\bf{1}}\\{ - {\bf{4}}}\end{array}} \right),\,\left( {\begin{array}{*{20}{c}}{ - {\bf{2}}}\\{\bf{9}}\end{array}} \right)\,} \right\}\). Since the coordinate mapping determined by B is a linear transformation from \({\mathbb{R}^{\bf{2}}}\) into \({\mathbb{R}^{\bf{2}}}\), this mapping must be implemented by some \({\bf{2}} \times {\bf{2}}\) matrix A. Find it. (Hint: Multiplication by A should transform a vector x into its coordinate vector \({\left( {\bf{x}} \right)_B}\)).

Question: Exercises 12-17 develop properties of rank that are sometimes needed in applications. Assume the matrix \(A\) is \(m \times n\).

15. Let \(A\) be an \(m \times n\) matrix, and let \(B\) be a \(n \times p\) matrix such that \(AB = 0\). Show that \({\mathop{\rm rank}\nolimits} A + {\mathop{\rm rank}\nolimits} B \le n\). (Hint: One of the four subspaces \({\mathop{\rm Nul}\nolimits} A\), \({\mathop{\rm Col}\nolimits} A,\,{\mathop{\rm Nul}\nolimits} B\), and \({\mathop{\rm Col}\nolimits} B\) is contained in one of the other three subspaces.)

Explain what is wrong with the following discussion: Let \({\bf{f}}\left( t \right) = {\bf{3}} + t\) and \({\bf{g}}\left( t \right) = {\bf{3}}t + {t^{\bf{2}}}\), and note that \({\bf{g}}\left( t \right) = t{\bf{f}}\left( t \right)\). Then, \(\left\{ {{\bf{f}},{\bf{g}}} \right\}\) is linearly dependent because g is a multiple of f.

Given \(T:V \to W\) as in Exercise 35, and given a subspace \(Z\) of \(W\), let \(U\) be the set of all \({\mathop{\rm x}\nolimits} \) in \(V\) such that \(T\left( {\mathop{\rm x}\nolimits} \right)\) is in \(Z\). Show that \(U\) is a subspace of \(V\).

If the null space of an \({\bf{8}} \times {\bf{5}}\) matrix A is 2-dimensional, what is the dimension of the row space of A?

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