Calculus: Analytic Geometry and Calculus, with Vectors

8.5 Integration by substitutingz = tan 012 477

7 It is easy to evaluate the integral

Kl = fx sin x dx

with the aid of the formula

fu(x)v'(x) dx = u(x)v(x)- fv(x)u'(x) dx

for integration by parts. Do it by filling in the first and thenthe second row
of the formulas
u(x) = v'(x) _
u'(x) = v(x) =
in the most useful way. Check the result by differentiation.
8 Do not forget our guiding principle. If you have an idea other than inte-
gration by parts for evaluating the integral

K2 = fx2 sin x dx,

write the author a letter. Otherwise, develop a strong heart by solving the

problem by integration by parts.

9 When we get good basic ideas we can understand and sometimes even

originate modifications of them. If Mr. Watson asks us to evaluate the integral

L

r^1

= J x-f-1-f-1dx'

we could eliminate the radical by setting x = tang 0, but the result is unlovely.

We can try to simplify matters by making a substitutionx = u(t) so cleverly

devised that x -j- 1 = t. The denominator will then be simply t + 1 and the

other details may be of such a nature that we can find L. Show that

L=2 x 1-2log( x-+I -1)+c.

Hint: The simple identity

t t+1-1 1

_7+_1_-f- 1 t -}- 1

enables us to integrate the first member in a hurry.

8.5 Integration by substituting z = tan j This section provides

useful experience in changing variables by treating a particular substitu-

tion which is sometimes useful, but the section can be omitted without

damaging understanding of the remainder of the book.

We begin by deriving formulas which enable us to use the substitution

(or change of variable) z = tan

2

to convert integrals of quotients of

polynomials in sin 0, cos 0, tan 0, cot 0, sec 0, and csc 0 into integrals