140. Transcendental Functions.

Owing to the immense variety of the different classes of transcendental functions, the theory of their integration is a good deal less systematic than that of the integration of rational or algebraical functions. We shall consider in order a few classes of transcendental functions whose integrals can always be found.

 

141. Polynomials in cosines and sines of multiples of \(x\).

We can always integrate any function which is the sum of a finite number of terms such as \[A\cos^{m} ax \sin^{m’} ax \cos^{n} bx \sin^{n’} bx\dots,\] where \(m\)\(m’\), \(n\)\(n’\), … are positive integers and \(a\)\(b\), … any real numbers whatever. For such a term can be expressed as the sum of a finite number of terms of the types \[\alpha\cos\{(pa + qb + \dots)x\},\quad \beta \sin\{(pa + qb + \dots)x\}\] and the integrals of these terms can be written down at once.

Example LI

1. Integrate \(\sin^{3} x \cos^{2} 2x\). In this case we use the formulae \[\sin^{3} x = \tfrac{1}{4}(3\sin x – \sin 3x),\quad \cos^{2} 2x = \tfrac{1}{2}(1 + \cos 4x).\] Multiplying these two expressions and replacing \(\sin x\cos 4x\), for example, by \(\frac{1}{2}(\sin 5x – \sin 3x)\), we obtain \[\begin{gathered} \tfrac{1}{16}\int (7\sin x – 5\sin 3x + 3\sin 5x – \sin 7x)\, dx\\ = – \tfrac{7}{16}\cos x + \tfrac{5}{48}\cos 3x – \tfrac{3}{80}\cos 5x + \tfrac{1}{112}\cos 7x.\end{gathered}\]

The integral may of course be obtained in different forms by different methods. For example \[\int \sin^{3}x \cos^{2}2x\, dx = \int(4\cos^{4}x – 4\cos^{2}x + 1) (1 – \cos^{2} x)\sin x\, dx,\] which reduces, on making the substitution \(\cos x = t\), to \[\int(4t^{6} – 8t^{4} + 5t^{2} – 1)\, dt = \tfrac{4}{7}\cos^{7} x – \tfrac{8}{5}\cos^{5}x + \tfrac{5}{3}\cos^{3}x – \cos x.\] It may be verified that this expression and that obtained above differ only by a constant.

2. Integrate by any method \(\cos ax \cos bx\), \(\sin ax \sin bx\), \(\cos ax \sin bx\), \(\cos^{2}x\), \(\sin^{3}x\), \(\cos^{4}x\), \(\cos x \cos 2x \cos 3x\), \(\cos^{3}2x \sin^{2}3x\), \(\cos^{5}x \sin^{7}x\). [In cases of this kind it is sometimes convenient to use a formula of reduction (Misc. Ex. 39).]

 

142. The integrals \(\int x^{n}\cos x\, dx\), \(\int x^{n}\sin x\, dx\) and associated integrals.

The method of integration by parts enables us to generalise the preceding results. For \[\begin{aligned}  \int x^{n}\cos x\, dx &= & &x^{n}\sin x &&- n\int x^{n-1}\sin x\, dx,\\ \int x^{n}\sin x\, dx &= &-&x^{n}\cos x &&+ n\int x^{n-1}\cos x\, dx,\end{aligned}\] and clearly the integrals can be calculated completely by a repetition of this process whenever \(n\) is a positive integer. It follows that we can always calculate \(\int x^{n}\cos ax\, dx\) and \(\int x^{n}\sin ax\, dx\) if \(n\) is a positive integer; and so, by a process similar to that of the preceding paragraph, we can calculate \[\int P(x, \cos ax, \sin ax, \cos bx, \sin bx, \dots)\, dx,\] where \(P\) is any polynomial.

Example LII

1. Integrate \(x\sin x\), \(x^{2}\cos x\), \(x^{2}\cos^{2}x\), \(x^{2}\sin^{2}x \sin^{2} 2x\), \(x\sin^{2}x \cos^{4}x\), \(x^{3}\sin^{3}\frac{1}{3}x\).

2. Find polynomials \(P\) and \(Q\) such that \[\int\{(3x – 1)\cos x + (1 – 2x)\sin x\}\, dx = P\cos x + Q\sin x.\]

3. Prove that \(\int x^{n}\cos x\, dx = P_{n}\cos x + Q_{n}\sin x\), where \[P_{n} = nx^{n-1} – n(n – 1)(n – 2) x^{n-3} + \dots,\quad Q_{n} = x^{n} – n(n – 1) x^{n-2} + \dots.\]

 

143. Rational Functions of \(\cos x\) and \(\sin x\).

The integral of any rational function of \(\cos x\) and \(\sin x\) may be calculated by the substitution \(\tan \frac{1}{2}x = t\). For \[\cos x = \frac{1 – t^{2}}{1 + t^{2}},\quad \sin x = \frac{2t}{1 + t^{2}},\quad \frac{dx}{dt} = \frac{2}{1 + t^{2}},\] so that the substitution reduces the integral to that of a rational function of \(t\).

Example LIII

1. Prove that \[\int \sec x\, dx = \log |\sec x + \tan x|,\quad \int \csc x\, dx = \log |\tan \tfrac{1}{2}x|.\]

[Another form of the first integral is \(\log |\tan(\frac{1}{4}\pi + \frac{1}{2}x)|\); a third form is \(\frac{1}{2}\log |(1 + \sin x)/(1 – \sin x)|\).]

2. \(\int \tan x\, dx = -\log |\cos x|\), \(\int \cot x\, dx = \log |\sin x|\), \(\int\sec^{2} x\, dx = \tan x\), \(\int \csc^{2} x\, dx = -\cot x\), \(\int \tan x\sec x\, dx = \sec x\), \(\int \cot x \csc x\, dx = -\csc x\).

[These integrals are included in the general form, but there is no need to use a substitution, as the results follow at once from § 119 and equation (5) of § 130.]

3. Show that the integral of \(1/(a + b\cos x)\), where \(a + b\) is positive, may be expressed in one or other of the forms \[\frac{2}{\sqrt{a^{2} – b^{2}}} \arctan \left\{t\sqrt{\frac{a – b}{a + b}}\right\},\quad \frac{1}{\sqrt{b^{2} – a^{2}}} \log \left|\frac{\sqrt{b + a} + t\sqrt{b – a}} {\sqrt{b + a} – t\sqrt{b – a}}\right|,\] where \(t = \tan\frac{1}{2}x\), according as \(a^{2} > b^{2}\) or \(a^{2} < b^{2}\). If \(a^{2} = b^{2}\) then the integral reduces to a constant multiple of that of \(\sec^{2}\frac{1}{2}x\) or \(\csc^{2}\frac{1}{2}x\), and its value may at once be written down. Deduce the forms of the integral when \(a + b\) is negative.

4. Show that if \(y\) is defined in terms of \(x\) by means of the equation \[(a + b\cos x)(a – b\cos y) = a^{2} – b^{2},\] where \(a\) is positive and \(a^{2} > b^{2}\), then as \(x\) varies from \(0\) to \(\pi\) one value of \(y\) also varies from \(0\) to \(\pi\). Show also that \[\sin x = \frac{\sqrt{a^{2} – b^{2}} \sin y}{a – b\cos y},\quad \frac{\sin x}{a + b\cos x}\, \frac{dx}{dy} = \frac{\sin y}{a – b\cos y};\] and deduce that if \(0 < x < \pi\) then \[\int \frac{dx}{a + b\cos x} = \frac{1}{\sqrt{a^{2} – b^{2}}} \arccos \left(\frac{a\cos x + b}{a + b\cos x}\right).\]

Show that this result agrees with that of Ex. 3.

5. Show how to integrate \(1/(a + b\cos x + c\sin x)\). [Express \(b\cos x + c\sin x\) in the form \(\sqrt{b^{2} + c^{2}} \cos(x – \alpha)\).]

6. Integrate \((a + b\cos x + c\sin x)/(\alpha + \beta\cos x + \gamma\sin x)\).

[Determine \(\lambda\)\(\mu\)\(\nu\) so that \[a + b\cos x + c\sin x = \lambda + \mu(\alpha + \beta\cos x + \gamma\sin x) + \nu(-\beta\sin x + \gamma\cos x).\] Then the integral is \[\mu x + \nu \log |\alpha + \beta\cos x + \gamma\sin x| + \lambda \int \frac{dx}{\alpha + \beta\cos x + \gamma\sin x}.]\]

7. Integrate \(1/(a\cos^{2} x + 2b\cos x\sin x + c\sin^{2} x)\). [The subject of integration may be expressed in the form \(1/(A + B\cos 2x + C\sin 2x)\), where \(A = \frac{1}{2}(a + c)\), \(B = \frac{1}{2}(a – c)\), \(C = b\): but the integral may be calculated more simply by putting \(\tan x = t\), when we obtain \[\int \frac{\sec^{2} x\, dx}{a + 2b\tan x + c\tan^{2} x} = \int \frac{dt}{a + 2bt + ct^{2}}.]\]

 

144. Integrals involving \(\arcsin x\), \(\arctan x\), and \(\log x\).

The integrals of the inverse sine and tangent and of the logarithm can easily be calculated by integration by parts. Thus \[\begin{aligned} \int \arcsin x\, dx &= x\arcsin x – \int \frac{x\, dx}{\sqrt{1 – x^{2}}} = x\arcsin x + \sqrt{1 – x^{2}},\\ \int \arctan x\, dx &= x\arctan x – \int \frac{x\, dx}{1 + x^{2}} = x\arctan x – \tfrac{1}{2} \log(1 + x^{2}),\\ \int \log x\, dx &= x\log x – \int dx = x(\log x – 1).\end{aligned}\]

It is easy to see that if we can find the integral of \(y = f(x)\) then we can always find that of \(x = \phi(y)\), where \(\phi\) is the function inverse to \(f\). For on making the substitution \(y = f(x)\) we obtain \[\int \phi(y)\, dy = \int xf'(x)\, dx = xf(x) – \int f(x)\, dx.\] The reader should evaluate the integrals of \(\arcsin y\) and \(\arctan y\) in this way.

Integrals of the form \[\int P(x, \arcsin x)\, dx,\quad \int P(x, \log x)\, dx,\] where \(P\) is a polynomial, can always be calculated. Take the first form, for example. We have to calculate a number of integrals of the type \(\int x^{m} (\arcsin x)^{n}\, dx\). Making the substitution \(x = \sin y\), we obtain \(\int y^{n}\sin^{m}y \cos y\, dy\), which can be found by the method of § 142. In the case of the second form we have to calculate a number of integrals of the type \(\int x^{m} (\log x)^{n}\, dx\). Integrating by parts we obtain \[\int x^{m}(\log x)^{n}\, dx = \frac{x^{m+1} (\log x)^{n}}{m + 1} – \frac{n}{m + 1} \int x^{m}(\log x)^{n-1}\, dx,\] and it is evident that by repeating this process often enough we shall always arrive finally at the complete value of the integral.


$\leftarrow$ 132-139. Integration of algebraical functions. Integration by rationalisation. Integration by parts Main Page 145. Areas of plane curves $\rightarrow$
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