INTEGRATION BY SUBSTITUTION
5 minute review. Recall the chain rule for differentiating a function of a function,
df du
d
namely dx
f (u(x)) = du
dx , then demonstrate and ask for suggestions on integration
by substitution using the warm-up examples below, running through these steps: (1)
dx
du
introduce a new variable u = u(x) (or x = x(u)); (2) replace the measure using dx = du
1
(or dx = du du); (3) integrate with respect to u; (4) replace u with x in the answer.
dx
Class warm-up. Integrate (a)
R
x2 exp(x3 )dx, (b)
R
x2
1+x3 dx
and/or (c)
R
√ dx .
1−x2
Problems. (Choose from the below)
1. Simple substitutions. Find the following indefinite integrals by choosing a
suitable substitution.
Z
Z
x
(e)
(a)
t cos(t2 − 1) dt
dx
2+1
x
Z
Z
3x
dx
(b)
sin2 x cos x dx
(f)
2 + a2
x
Z
Z p
cos t
√
(g)
dt
(c)
x 1 + x2 dx
1 + sin t
Z
Z
p
cosh u
(d) (t + 1) t2 + 2t dt
(h)
du
1 + sinh2 u
2. Trigonometric substitutions. Find the indefinite integrals using the suggested
substitutions.
Z
Z
1
1
(a)
dx,
3x
=
tan
u;
(d)
dx, x + 1 = 2 tan u;
2
2 + 2x + 5
1
+
9x
x
Z
Z
1
√
(b)
dx, 4x = 5 sin u; (e)
sec 2x tan 2x dx, u = cos 2x.
25 − 16x2
Z
(c)
tan x dx,
u = cos x;
3. Completing the square. Find the following.
Z
Z
dx
cos x dx
(c)
(a)
2
2 + 6x + 10
x
Z
Z sin x + 2 sin x + 10
dx
dx
√
(b)
(d)
.
x2 + 10x + 29
3 − 2x − x2
4. Log-of-a-log? . Use the substitution x = eu to show that
Z
dx
= ln (ln x) + c.
x ln x
Let In (x) be defined as follows.
Z
Z
Z
dx
dx
dx
I1 (x) =
, I2 (x) =
, I3 (x) =
,...
x ln x
x ln(x) ln(ln x)
x ln(x) ln(ln x) ln(ln(ln x))
Note that, as n gets larger, x has to be very large and positive in order that its
logarithm can be taken repeatedly.
Find I2 (x) using integration by substitution. Show that In (eu ) = In−1 (u) (in
other words, that substituting x = eu in In (x) gives In−1 (u)), and hence write
down a general formula for In (x). Check it is valid by differentiating.
INTEGRATION BY SUBSTITUTION
R
For the warm-up, x2 exp(x3 )dx =
R dx
√
= arcsin x.
1−x2
1
3
exp(x3 ), (b)
R
x2
1+x3 dx
=
1
3
ln(1 + x3 ) and (c)
Selected answers and hints. (All answers should include a constant of integration.)
3/2
1. (a) 12 sin(t2 −1), (b) 13 sin3 x, (c) 13 1 + x2
, (d) 13 (t2 +2t)3/2 (e) 21 ln(1+x2 ),
√
(f) 23 ln x2 + a2 , (g) 2 1 + sin t, (h) tan−1 (sinh u).
2. (a) 13 arctan(3x), (b) 14 arcsin 54 x , (c) ln | sec x|, (d) 12 tan−1 21 (x + 1) , (e)
1
2 sec(2x).
3. (a) tan−1 (x+3), (b) sin−1 21 (x + 1) , (c) 13 tan−1 31 (sin x + 1) (d) 12 tan−1 12 (x + 5) .
4. In (eu ) is the result of making the substitution x = eu in In (x). That is,
Z
dx
In (eu ) =
u
u
e ln(e ) . . . ln(ln . . . (ln eu ) . . .)
Z
dx
du du
=
u
e u . . . ln(ln . . . (u) . . .)
Z
eu du
=
u
e u . . . ln(ln . . . (u) . . .)
= In−1 (u).
Putting x = ln u in the given identity, we get
In (x) = In−1 (ln x) = In−2 (ln(ln x)) = . . .
It follows that
In (x) = ln(ln(ln . . . (ln(x)) . . .)) +c.
|
{z
}
n+1 times
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