## 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.