An introduction to the study of the elements of the differential and integral calculus. From the German of the late Axel Harnack, With the permission of the author.

238

Integrals of transcendental functions.

Bk. III. ch. V.

For n = 0 we had already:
re d ekX
jekdx --.
Every integral of this kind, in which n is a positive integer, is reduced
by formula 3. to one of these two integrals.
When k is a negative number, the integral can be taken up to
a positive infinite limit, for although sin x and cos x become quite
indeterminate between - 1 and + 1, yet ekx in the integral function
will pass over continuously into zero. Thus we have for k < 0:
ekxsinxdx = 72
0
In like manner we find:
4. fl xd e cos' x (k cos x + sin x
4. kcosnX x + 2
n(n - COS1) -2X ( X
Here
fex cosxdx ekx (k cosx + sin x)
eX cosxdx 2 L - -- -
and for k < 0:
k2 + 1
0
141. If circular functions occur in the function to be integrated,
the process of integration by parts leads likewise in many cases to a
solution or simplification of the problem. If in the integral
fX sin-~x ~ dx
the function X be integrable we hav.e:
sin- ~xdx = — sin-1 x Xdx - J 1 X dx
Ex.gr.: fxnsi 7 +x 1 x sin dx
Ex. gr.: I sin-' X X -- sin- - X -- --
n+ 1 ~n+ I yh x2
This new binomial integral can be expressed in finite terms when n
is an integer. We can also get rid of the circular functions by
introducing algebraic and trigonometric functions, putting
sin-1x =z, x=sin, dx = cos zdz.
Thus ex. gr.: J(sin-l x) X x = n coszdz.