Calculus: Analytic Geometry and Calculus, with Vectors

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8.4 Integration by trigonometric andother substitutions 475

Problems 8.49


1 By making an appropriate figure and trigonometric substitution,fill in all
of the missing details which show that

(a) r x21-a2dx = r sec 9 dB = log (x -f x2 - a2) + c

(b) !(a2 x2dx = a f sec 8 d9

=flog a+x +c logo+x+c
a a2-,x2 2a a-x (JxI <a)

(c) f

x2 a2

dx =
Q

f csc 8 dB

(^1) l x+ a (^1) x- a



  • C
    a
    log
    x2 --a 2


+ 2alegx + a

(d) f (a2 - x2)3 dx = a° f cos' 0 dO

2 When x > 0, the identity

(1)

1 + 1 = z2^1 - x2 + 1
x x x 1 + X2
shows that

(2) f 1 +^1 dx = f (1 +


x x 1+x2


(x > a)

The first integral on the right is easily evaluated, and the second can be simplified
by a trigonometric substitution. Use these ideas to derive the formula

(3) f rl ± .2dx = 1 x2 + log x - log (1 + 1 -+x2) + c.


3 Sketch graphs of y = ex and y = log x in a single figure and note that each
is the mirror image of the other in the line y = x. Observe that the arc C, on
the graph of y = ex which joins two points (p,,q,) and (p2,Q2) on the graph of
y = ex is congruent to the arc C2 on the graph of y = log x which joins the two
points (q,,p,) and (g2,p2) on the graph of y = log x. LettingL denote the length
of C, and C2, show that


(1)

L =

f


p:
1 e2x dx = f 'D'V+1 dx.
P, 'D, x2
This gives the formula
(2)
J

1 + eit dt =

I.
`y' 1 + 1 dx.
P

1
i en2 x2

To show that errors and misprints have not led us to an incorrect formula, show
that the substitution i = log x converts the first integral in (2) into the second
integral in (2).

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