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 Acosmaxsinmaxcosnbxsinnbx, where mm, nn, … are positive integers and ab, … any real numbers whatever. For such a term can be expressed as the sum of a finite number of terms of the types αcos{(pa+qb+)x},βsin{(pa+qb+)x} and the integrals of these terms can be written down at once.

Example LI

1. Integrate sin3xcos22x. In this case we use the formulae sin3x=14(3sinxsin3x),cos22x=12(1+cos4x). Multiplying these two expressions and replacing sinxcos4x, for example, by 12(sin5xsin3x), we obtain 116(7sinx5sin3x+3sin5xsin7x)dx=716cosx+548cos3x380cos5x+1112cos7x.

The integral may of course be obtained in different forms by different methods. For example sin3xcos22xdx=(4cos4x4cos2x+1)(1cos2x)sinxdx, which reduces, on making the substitution cosx=t, to (4t68t4+5t21)dt=47cos7x85cos5x+53cos3xcosx. It may be verified that this expression and that obtained above differ only by a constant.

2. Integrate by any method cosaxcosbx, sinaxsinbx, cosaxsinbx, cos2x, sin3x, cos4x, cosxcos2xcos3x, cos32xsin23x, cos5xsin7x. [In cases of this kind it is sometimes convenient to use a formula of reduction (Misc. Ex. 39).]

 

142. The integrals xncosxdx, xnsinxdx and associated integrals.

The method of integration by parts enables us to generalise the preceding results. For xncosxdx=xnsinxnxn1sinxdx,xnsinxdx=xncosx+nxn1cosxdx, 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 xncosaxdx and xnsinaxdx if n is a positive integer; and so, by a process similar to that of the preceding paragraph, we can calculate P(x,cosax,sinax,cosbx,sinbx,)dx, where P is any polynomial.

Example LII

1. Integrate xsinx, x2cosx, x2cos2x, x2sin2xsin22x, xsin2xcos4x, x3sin313x.

2. Find polynomials P and Q such that {(3x1)cosx+(12x)sinx}dx=Pcosx+Qsinx.

3. Prove that xncosxdx=Pncosx+Qnsinx, where Pn=nxn1n(n1)(n2)xn3+,Qn=xnn(n1)xn2+.

 

143. Rational Functions of cosx and sinx.

The integral of any rational function of cosx and sinx may be calculated by the substitution tan12x=t. For cosx=1t21+t2,sinx=2t1+t2,dxdt=21+t2, so that the substitution reduces the integral to that of a rational function of t.

Example LIII

1. Prove that secxdx=log|secx+tanx|,cscxdx=log|tan12x|.

[Another form of the first integral is log|tan(14π+12x)|; a third form is 12log|(1+sinx)/(1sinx)|.]

2. tanxdx=log|cosx|, cotxdx=log|sinx|, sec2xdx=tanx, csc2xdx=cotx, tanxsecxdx=secx, cotxcscxdx=cscx.

[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+bcosx), where a+b is positive, may be expressed in one or other of the forms 2a2b2arctan{taba+b},1b2a2log|b+a+tbab+atba|, where t=tan12x, according as a2>b2 or a2<b2. If a2=b2 then the integral reduces to a constant multiple of that of sec212x or csc212x, 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+bcosx)(abcosy)=a2b2, where a is positive and a2>b2, then as x varies from 0 to π one value of y also varies from 0 to π. Show also that sinx=a2b2sinyabcosy,sinxa+bcosxdxdy=sinyabcosy; and deduce that if 0<x<π then dxa+bcosx=1a2b2arccos(acosx+ba+bcosx).

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

5. Show how to integrate 1/(a+bcosx+csinx). [Express bcosx+csinx in the form b2+c2cos(xα).]

6. Integrate (a+bcosx+csinx)/(α+βcosx+γsinx).

[Determine λμν so that a+bcosx+csinx=λ+μ(α+βcosx+γsinx)+ν(βsinx+γcosx). Then the integral is μx+νlog|α+βcosx+γsinx|+λdxα+βcosx+γsinx.]

7. Integrate 1/(acos2x+2bcosxsinx+csin2x). [The subject of integration may be expressed in the form 1/(A+Bcos2x+Csin2x), where A=12(a+c), B=12(ac), C=b: but the integral may be calculated more simply by putting tanx=t, when we obtain sec2xdxa+2btanx+ctan2x=dta+2bt+ct2.]

 

144. Integrals involving arcsinx, arctanx, and logx.

The integrals of the inverse sine and tangent and of the logarithm can easily be calculated by integration by parts. Thus arcsinxdx=xarcsinxxdx1x2=xarcsinx+1x2,arctanxdx=xarctanxxdx1+x2=xarctanx12log(1+x2),logxdx=xlogxdx=x(logx1).

It is easy to see that if we can find the integral of y=f(x) then we can always find that of x=ϕ(y), where ϕ is the function inverse to f. For on making the substitution y=f(x) we obtain ϕ(y)dy=xf(x)dx=xf(x)f(x)dx. The reader should evaluate the integrals of arcsiny and arctany in this way.

Integrals of the form P(x,arcsinx)dx,P(x,logx)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 xm(arcsinx)ndx. Making the substitution x=siny, we obtain ynsinmycosydy, 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 xm(logx)ndx. Integrating by parts we obtain xm(logx)ndx=xm+1(logx)nm+1nm+1xm(logx)n1dx, and it is evident that by repeating this process often enough we shall always arrive finally at the complete value of the integral.


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