Download MAT 136F Tutorial 4 – Integration methods II October 26–30, 2015

MAT 136F
Tutorial 4 – Integration methods II
October 26–30, 2015
1. Evaluate the following integrals.
Z
−3x + 17
(a)
dx
x2 − 2x − 3
Z
4 − 4x
(b)
dx
2
x − 2x − 3
Hint: This is easier than the
previous one.
Z
x2
√
(c)
dx
9 − x2
Z
x
√
(d)
dx
9 − x2
Hint: This is easier than the
previous one.
Z
x
√
dx
5 − 4x − x2
Z
3x2 + 2x + 1
dx
x3 + x2 + x
Z
4x4 − x2 + 6x − 1
dx
4x2 − 1
(e)
(f)
(g)
Z
(h)
ex
Z
(i)
√
dx
dx
4 + e2x
dx
√
4x2 + 2x
2. The integral of sec x. In your textbook, the integral of sec x was calculated using some
trickery. Here we will study another method for integrating sec x, that is longer, but less
dependent on a memorized trick.
(a) Convince yourself that sec x =
cos x
.
1 − sin2 x
Z
(b) Try a substitution to transform the integral
cos x
dx into the integral of a
1 − sin2 x
rational function.
(c) Use partial fractions to evaluate the integral you obtained in part (b).
(d) Write your answer in terms of the original variable, x.
Z
1 sin x + 1 Your final answer should look like sec x dx = ln +C.
2
sin x − 1 Notice that this looks different from the answer in your textbook. However, with a little
bit of work, it is possible to see that the answer that you got is really the same.
1 sin x + 1 (e*) (optional) Show that ln = ln sec x + tan x
2
sin x − 1
Hint: Start with the left-hand side, multiply
√
Next, remember that 21 ln a = ln a.
sin x+1
sin x−1
by
sin x+1
.
sin x+1
MAT 136F
Tutorial 4 – Integration methods II
October 26–30, 2015
Solutions
−5
2
−3x + 17
=
+
1. (a) Use partial fractions to get: 2
x − 2x − 3
x+1 x−3
Z
−3x + 17
Answer:
dx = −5 ln |x + 1| + 2 ln |x − 3| + C
x2 − 2x − 3
(b) Substitution u = x2 − 2x − 3
Z
4 − 4x
Answer:
dx = −2 ln |x2 − 2x − 3| + C
2
x − 2x − 3
(c) Trigonometric Substitution: x = 3 sin θ
Z
x2
9
x x√
√
Answer:
dx = sin−1 −
9 − x2 + C
2
2
3
2
9−x
(d) Substitution: u = 9 − x2
Z
x
√
Answer:
dx = −(9 − x2 )1/2 + C
2
9−x
x
x
. Then use the trigono=p
5 − 4x − x2
9 − (x + 2)2
metric substitution x = 3 sin θ − 2.
Z
√
x
x+2
−1
2
√
Answer:
dx = − 5 − 4x − x − 2 sin
+C
3
5 − 4x − x2
(e) First complete the square: √
3x2 + 2x + 1
1
2x + 1
(f) Use partial fraction to obtain: 3
= + 2
.
2
x +x +x
x x +x+1
Z
3x2 + 2x + 1
Answer:
dx = ln |x| + ln |x2 + x + 1| + C
x3 + x2 + x
4x4 − x2 + 6x − 1
1
2
(g) Use long division and partial fractions to get:
= x2 +
+
.
2
4x − 1
2x − 1 2x + 1
Z
4x4 − x2 + 6x − 1
x3 1
dx
=
+ ln |2x − 1| + ln |2x + 1| + C
Answer:
4x2 − 1
3
2
(h) Substitution ex = 2 tan θ. Or, if you prefer, you can do it in two steps: first ex = u,
then do a trig substitution.
√
Z
dx
4 + e2x
√
Answer:
=−
+C
4ex
ex 4 + e2x
1
1
(i) Complete the square. Then trig substitution 2x + = sec θ, which is the same as
2
2
sec θ = 4x + 1.
Z
i
√
dx
1 h
√
Answer:
= ln 4x + 1 + 2 4x2 + 2x + C
2
4x2 + 2x
1
, multiplying by cos x to both the numerator and denominator,
2. (a) Because sec x =
cos x
we obtain that
cos x
cos x
=
sec x =
.
2
cos x
1 − sin2 x
(b) Let u = sin x. Then
Z
cos x
dx =
1 − sin2 x
Z
1
du.
1 − u2
(c)
Z
1
du =
1 − u2
Z
1
2
1
1
+
1+u 1−u
du =
1
(ln |1 + u| − ln |1 − u|) + C.
2
(d) Since u = sin x,
1
1 sin x + 1 (ln |1 + u| − ln |1 − u|) + C = ln + C.
2
2
sin x − 1 Now consider
sin x + 1 (sin x + 1)2
=
sin x − 1 (sin x − 1)(sin x + 1) .
And by applying the trigonometric identity sin2 x + cos2 x = 1, what can you obtain?
MAT 136F
Tutorial 4 – Integration methods II
October 26–30, 2015
Detailed solutions for selected questions
1. (a) Notice that the denominator x2 − 2x − 3 = (x + 1)(x − 3). We may assume the
integrand is the sum of the partial fractions:
A
B
−3x + 17
=
+
,
2
x − 2x − 3
x+1 x−3
where A and B are constants to be determined. Multiplying by the least common
denominator, we get
A(x − 3) + B(x + 1)
(A + B)x − 3A + B
−3x + 17
=
=
.
2
2
x − 2x − 3
x − 2x − 3
x2 − 2x − 3
Comparing the numerators, we obtain the following system of equations for A and
B:
A+B
= −3
−3A + B = 17
By solving the system, we get A = −5 and B = 2, and therefore,
Z Z
−5
2
−3x + 17
dx =
+
dx
x2 − 2x − 3
x+1 x−3
= −5 ln |x + 1| + 2 ln |x − 3| + C
(c) Consider the trigonometric substitution: x = 3 sin θ. Then dx = 3 cos θdθ. And
Z
Z
9 sin2 θ
x2
√
p
dx =
· 3 cos θ dθ
9 − x2
9 − 9 sin2 θ
Z
Z
9 sin2 θ cos θ
=
dθ = 9 sin2 θ dθ
cos θ
Z
1
=9
(1 − cos 2θ) dθ
2
9θ 9 sin 2θ
=
−
+ C.
2
4
Now notice that because x = 3 sin θ, we have θ = sin−1 x3 . Then
x
sin 2θ = 2 sin θ cos θ = 2 · ·
3
r
1−
x2
2 √
= x 9 − x2 .
9
9
Therefore, we obtain that
Z
x x√
x2
9
√
dx = sin−1 −
9 − x2 + C.
2
2
3
2
9−x
(e) We first complete the square for the polynomial under the square root:
5 − 4x − x2 = 5 − (x2 + 2 · 2x + 4) + 4 = 9 − (x + 2)2 .
This suggests we make a trigonometric substitution x + 2 = 3 sin θ.
Then x = 3 sin θ − 2 and dx = 3 cos θdθ. So our integral becomes
Z
Z
Z
x
3 sin θ − 2
x
√
p
p
dx =
dx =
· 3 cos θ dθ
5 − 4x − x2
9 − (x + 2)2
9 − 9 sin2 θ
Z
= (3 sin θ − 2) dθ = −3 cos θ − 2θ + C.
p
and
cos
θ
=
1 − sin2 θ =
Now x + 2 = 3 sin θ implies that θ = sin−1 x+2
3
q
2
1 − x+2
. After simplifying the expression, we obtain that
3
Z
√
x
+
2
x
−1
√
dx = − 5 − 4x − x2 − 2 sin
+ C.
3
5 − 4x − x2
(h) We first substitute ex = u. Then du = ex dx, or
dx =
1
1
du = du.
x
e
u
The integral then becomes
Z
Z
Z
1
dx
1
1
√
√
√
=
· du =
du.
x
2x
2
2
e 4+e
u 4+u u
u 4 + u2
Now consider a substitution u = 2 tan θ. Then du = 2 sec2 θdθ, so
Z
Z
1
1
√
du =
· 2 sec2 θ dθ
2
2
2
4 tan θ · 2 sec θ
u 4+u
Z
Z
1
sec θ
1
cos θ
=
dθ =
dθ
2
4
tan θ
4
sin2 θ
1
=−
+ C.
4 sin θ
Because tan θ = u/2,
u
ex
sin θ = √
=√
.
u2 + 22
4 + e2x
Therefore,
√
Z
dx
4 + e2x
√
=−
+ C.
4ex
ex 4 + e2x