INTEGRATION BY SUBSTITUTION AND PARTS
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 explain integration, 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
Recap
perhaps with the example
R 1 how partial fractions can be used in integration,
d
dx.
Recall
the
product
rule
for
differentiation,
(uv)
= u0 v + uv 0 , and inte1−x2
dx
R 0
R
grate and rearrange to obtain the integration by parts formula uv dx = uv − u0 v dx.
R
R x2
R
4
Class warm-up.
Integrate (a) x2 exp(x3 )dx, (b) 1+x
3 dx, (c)
(x−1)(x+1)2 dx
R
(using partial fractions), (d) x ln x dx (using integration by parts).
Problems. (Choose from the below)
I. Simple substitutions. Find the following indefinite integrals by choosing a suitable substitution.
Z
Z
x
(a)
t cos(t2 − 1) dt
dx
(c)
2+1
x
Z p
Z
cos t
√
(b)
x 1 + x2 dx
dt
(d)
1 + sin t
II. Trigonometric substitutions. Find the indefinite integrals using the suggested
substitutions.
Z
Z
1
1
√
(a)
dx, 4x = 5 sin u
dx,
3x
=
tan
u
(b)
2
1 + 9x
25 − 16x2
III. Completing the square. Find the following.
Z
Z
dx
dx
√
(a)
(b)
x2 + 6x + 10
3 − 2x − x2
IV. Partial fractions. Find the following indefinite integrals using the method of
partial fractions as appropriate.
Z
Z
(x + 1)
y dy
(a)
dx
(c)
3
2
Z y −y +y−1
Z x(x + 3)
3x + 3
dx
(d)
dx
(b)
(x − 1)3 (2x + 1)
(x2 + 1)(x + 1)
V. Integration by parts. Evaluate the following using integration by parts.
Z
Z
Z
Z
2
(a) tet dt; (b) ln x dx; (c) y 3 e−y dy; (d) cosh−1 u du.
VI. 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 AND PARTS
For the warm-up: (a)
R
x2 exp(x3 )dx =
1
3
1
1
2
4
2 = x−1 − x+1 − (x+1)2 , so
R(x−1)(x+1)
(d) x ln x dx = 14 x2 (2 ln x − 1).
(c)
exp(x3 ), (b)
R
4 dx
(x−1)(x+1)2
x2
1
3
1+x3 dx = 3 ln(1 + x ),
x−1 2
= ln x+1 + x+1
+ c, and
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Selected answers and hints. (All answers should include a constant of integration.)
3/2
√
, (c) 21 ln(1 + x2 ), (d) 2 1 + sin t.
I. (a) 12 sin(t2 − 1), (b) 13 1 + x2
II. (a) 13 arctan(3x), (b) 14 arcsin 45 x .
III. (a) tan−1 (x + 3), (b) sin−1 12 (x + 1) .
1
1
−1
2
IV. (a) 13 ln x(x + 3)2 , (b) 12 ln |x + 1|− 14 ln
|1 + x | + 2 tan (x) (c) 2 ln |y − 1| −
1
1
x−1
1
1
−1
2
(y), (d) 29 ln 2x+1
− (x−1)
+ 3(x−1)
2.
4 ln |1 + y | + 2 tan
V. (a) (t − 1)et , (b) x(ln x − 1) using u = ln x, v 0 = 1, (c) − 12 y 2 + 1 exp(−y 2 ),
√
(d) u cosh−1 u − u2 − 1.
VI. In (eu ) is the result of making the substitution x = eu in In (x). That is,
Z
dx
In (eu ) =
eu ln(eu ) . . . ln(ln . . . (ln eu ) . . .)
Z
dx
du du
=
eu 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
For more details, start a thread on the discussion board.
INTEGRATION BY SUBSTITUTION AND PARTS
Extra Problems.
I. Substitutions. Find the following indefinite integrals by choosing a suitable substitution.
Z
Z
3x
dx
(a)
sin2 x cos x dx
(c)
2
2
Z
Z x +a
p
cosh u
(b) (t + 1) t2 + 2t dt
(d)
du
1 + sinh2 u
II. Trigonometric substitutions. Find the indefinite integrals using the suggested
substitutions.
Z
Z
(a)
tan x dx, u = cos x;
(c)
sec 2x tan 2x dx, u = cos 2x.
Z
1
(b)
dx, x + 1 = 2 tan u;
x2 + 2x + 5
III. Completing the square. Find the following.
Z
Z
dx
cos x dx
(b)
(a)
2
2 + 10x + 29
x
sin x + 2 sin x + 10
IV. Partial fractions. Find the following indefinite integrals using the method of
partial fractions as appropriate.
Z
Z
dz
dx
(b)
(a)
x2 + 3x − 10
(z − 2)2 (z + 1)
V. Integration by parts. Evaluate the following using integration by parts.
Z
Z
(a)
x2 cosh x dx
(c)
ln(t2 + a2 ) dt
Z
Z
(b)
xn ln x dx (n 6= −1)
(d)
tan−1 u du
VI. Recurrence formulae? .
R
(a) Let In = xn eax dx where n ≥ 0 is an integer and a is a (possibly complex)
constant. Using integration by parts, show that, for n > 0,
1
In = (xn eax − nIn−1 )
a
and I0 = a1 eax + c. Find I1 , I2 and I3 . Show that
In
1 xn ax
In−1
=
e −
,
n!
a n!
(n − 1)!
and find a general expression for In .
R
R
(b) Let Cn = xn cos x dx and Sn = xn sin x dx. Show that, for n > 0,
xn sin x − nSn−1 ,
Cn
=
Sn
= −xn cos x + nCn−1 .
and C0 = sin x + c, S0 = − cos x + c. Following a similar approach to above,
can you find a general expression for Cn and Sn ?
(c) When a = i, what’s the relationship between In , Sn and Cn ?
INTEGRATION BY SUBSTITUTION AND PARTS
Selected answers and hints. (All answers should include a constant of integration.)
I. (a) 13 sin3 x, (b) 13 (t2 + 2t)3/2 (c) 32 ln x2 + a2 .
II. (a) ln | sec x|, (b) 21 tan−1 12 (x + 1) , (c) 12 sec(2x).
III. (a) 13 tan−1 13 (sin x + 1) (b) 21 tan−1 12 (x + 5) .
z+1 1
1
(b)
IV. (a) 17 ln x−2
ln
,
x+5
9
z−2 − 3(z−2) .
n+1
x
V. (a) (x2 + 2) sinh x − 2x cosh x, (b) (n+1)
2 ((n + 1) ln x − 1) ,
−1
2
2
(c) t ln(t + a ) − 2t + 2a tan (t/a), (d) u tan−1 u − 12 ln |1 + u2 |.
VI. (a) A general formula for In is
(ax)n
(ax)n−1
(ax)n−2
eax
n−1
n
In = n! n+1
−
+
− . . . + (−1)
ax + (−1)
+c
a
n!
(n − 1)!
(n − 2)!
n
eax X (−1)k (ax)n−k
+ c.
= n! n+1
a
(n − k)!
k=0
(b) The general formulas are
n
xn−2
xn−4
x
−
+
− . . . sin x
Cn = n!
n!
(n − 2)! (n − 4)!
n−1
x
xn−3
xn−5
+n!
−
+
− . . . cos x + c
(n − 1)! (n − 3)! (n − 5)!
and
n−1
xn−3
xn−5
x
−
+
− . . . sin x
Sn = n!
(n − 1)! (n − 3)! (n − 5)!
n
x
xn−2
xn−4
−n!
−
+
− . . . cos x + c
n!
(n − 2)! (n − 4)!
where the series in each bracket terminates before the power of x involved
becomes negative.
(c) Using Euler’s relation, eix = cos x + i sin x, we find that In = Cn + iSn .
For more details, start a thread on the discussion board.