Download Handout 6

Handout 6
Research shows that students learn when they MAKE MISTAKES and not when they get it right. SO MAKE MISTAKES,
because MISTAKES ARE GOOD!
1. Useful Integrals and Identities
∫
∫
tan(x)dx = ln | sec(x)| + C
sin2 (x) =
sec(x)dx = ln | sec(x) + tan(x)| + C
1
(1 − cos(2x))
2
cos2 (x) =
1
(1 + cos(2x))
2
sin2 (x) + cos2 (x) = 1
1 − sin2 (x) = cos2 (x)
1 + tan2 (x) = sec2 (x)
sec2 (x) − 1 = tan2 (x)
2. Trigonometric Substitution
∫
∫
Notes: The main idea to remember when encountering integrals of the form [sin(x)]n [cos(x)]m dx or [tan(x)]n [sec(x)]m dx,
n, m whole numbers, is to treat them like you treat integrals involving U-Substitution, because you will eventually be
using U-Substitution to solve them. The difference is that you might have to use the half angle formulas or the Fundamental Trig Identity beforehand. Knowing when is a matter of practice and familiarity.
∫
Ex.1
tan5 (x) sec(x)dx
When attempting U-Sub the only suitable options are u = tan(x) or u = sec(x) which means I need du = sec2 (x) or
sec(x) tan(x). There is no sec2 (x) but there is a sec(x) tan(x). You might wonder about the extra tangents but have
no fear use the trig ids.
∫
∫
tan5 (x) sec(x)dx =
tan4 (x) sec(x) tan(x)dx
∫
=
∫
=
∫
=
∫
=
[ 2 ]2
tan (x) sec(x) tan(x)dx
[ 2
]2
sec (x) − 1 sec(x) tan(x)dx
[ 2
]2
u − 1 du
(
when u = sec(x) and du = sec(x) tan(x)
)
u4 − 2u2 + 1 du
u5
2u3
−
+u+C
5
3
sec5 (x) 2 sec3 (x)
=
−
+ sec(x) + C
5
3
=
Notes: The main idea for the other types of Trig-Sub is knowing when to substitute
the variable for a trig function.
√
This will happen almost always when there is an integral with the expression ax2 + b in there somewhere for real
numbers a and b. The easiest way to do these problems is to memorize the three trig id’s above that you get from the
Fundamental Trig Identity and making the expression under the square root look like one of them.
1
∫
Ex.2
x3
√
x2 + 4
dx
The expression under the square root x2 + 4 looks most closely like the Trig Identity tan2 (x) + 1 = sec2 (x) so lets make
it look exactly like it then integrate as follows.
∫
√ [
]
x2
3
dx = x 4
+1
dx
4
√[
]
∫
( x )2
3
+1
dx
= x 2
2
∫
√[
] [
]
3
= 2 (2 tan(u))
tan2 (u) + 1 2 sec2 (u) du
if x = 2 tan(u),
∫
√
= 32 tan3 (u) sec2 (u) sec2 (u)du
∫
= 32 tan3 (u) sec3 (u)du
∫
= 32 [sec2 (u) − 1] tan(u) sec2 (u) sec(u)du
∫
= 32 [v 2 − 1]v 2 dv
if v = sec(u), dv = sec(u) tan(u)du
[ 5
]
v
v3
= 32
−
+C
5
3
[ 5
]
sec (u) sec3 (u)
= 32
−
+C
5
3
∫
√
x3 x2 + 4
√
[
]
dx = 2 sec2 (u) du
2
Since tan(u) = x2 then sec(u) = x2 +4 . To see this draw a right triangle with side opposite the angle u as x and side
adjacent to the angle u as 2. Then use the pythagorean thm. to find the hypotenuse.
(√

= 32 
x2 +4
2
5
)5
(√
−
x2 +4
2
3
)3 

+C
3. Partial Fractions
The main idea concerning partial fractions is to transform ”hard” rational integrals into the sum of integrals of ”easier”
rational function using the technique of Partial Fractions. Here is how the technique goes: Suppose P (x) is some
polynomial and you have a completely reduced and factored integral that looks like the following:
∫
P (x)
m
n dx
2
(ax + bx + c) (dx + e)
then you can find constants Aj , Bj and Ck , j ∈ [1, m], k ∈ [1, n] such that
∫
∑
P (x)
dx
=
(ax2 + bx + c)m (dx + e)n
i=1
m
∫
n ∫
∑
Ai x + Bi
Ci
dx +
dx
2
i
(ax + bx + c)
(dx + e)i
i=1
Abstractly since you know everything except the constants A1 ...Am , B1 ...Bm , C1 ...Cn you can solve for them by solving:
(ax2
P (x)
A2 x + B2
Am x + Bm
C1
Cn
A1 x + B1
+
+ ... +
+
+ ... +
m
n =
2
2
2
2
m
(ax + bx + c) (ax + bx + c)
(ax + bx + c)
(dx + e)
(dx + e)n
+ bx + c) (dx + e)
How do you know when to use Ax + B or when to use just a constant C? The answer is once you get the denominator
into its most factored form then you always put Ax + B above the factor of degree 2 and just a constant above the
factors of degree 1, the power the factor is being raised to tells you how many times you do it increasing the power of
the denominator by 1 each time (these were the m and n from above). In fact in general you just write the general
2
equation of 1 degree less then the irreducible factor below it, every time.
4. Examples:
a.)
∫1
x−4
dx
− 5x + 6
x2
0
First we factor the denominator as much as possible.
∫1
0
x−4
dx =
2
x − 5x + 6
∫1
0
x−4
dx
(x − 2)(x − 3)
Now we setup a partial fractions equality and solve by clearing the fractions as follows:
x−4
A
B
=
+
(x − 2)(x − 3)
(x − 2) (x − 3)
⇒ x − 4 = A(x − 3) + B(x − 2)
⇒ 1 ∗ x + (−4) = (A + B)x + (−3A − 2B)
⇒ 1 = (A + B) and − 4 = −3A − 2B
⇒ A = 1 − B → −4 = −3(1 − B) − 2B → B = −1,
A=2
Once we have our unkown variables we can split the original integral into two ”easy” ones and solve as follows:
∫1
⇒
0
x−4
=
(x − 2)(x − 3)
∫1 [
0
∫1
=2
0
]
−1
2
+
dx
(x − 2) (x − 3)
1
dx −
x−2
∫1
0
1
dx
x−3
1
1
= 2 [ln |x − 2|]0 − [ln |x − 3|]0
= 2 [0 − ln(2)] − [ln(2) − ln(3)]
= ln(3) − 3 ln(2)
5. Exercises:
(a)
∫1
0
x3 − 4x − 10
dx
x2 − x − 6
Setting up the partial fractions equality after we use long division and solving goes as follows:
2x − 10
x3 − 4x − 10
=x+1+ 2
x2 − x − 6
x −x−6
A
B
2x − 10
=
+
⇒ 2
x −x−6
x−3 x+2
⇒ 2x − 10 = A(x + 2) + B(x − 3)
3
By choosing x = −2 and x = 3 we can find A and B as
(x = 3) ⇒ −4 = A5
−4
⇒A=
5
(x = −2) ⇒ −14 = −5B
14
⇒B=
5
[
]
∫1
∫1 3
−4
x − 4x − 10
14
x
+
1
+
dx
=
+
dx
⇒
x2 − x − 6
5(x − 3) 5(x + 2)
0
0
[
]1
x2
−4
14
=
+x+
ln |x − 3| +
ln |x + 2|
2
5
5
0
1
−4
14
4
14
= +1+
ln(2) +
ln(3) + ln(3) −
ln(2)
2
5
5
5
5
(b)
∫ √
9 − x2
dx
x
Under the square root looks mostly like 1 − sin2 (θ) so we make it look like that as follows:
∫ √
∫ √
3 1 − ( x3 )2
9 − x2
dx =
dx
x
x
Let
x
3
= sin(θ) so that dx = 3 cos(θ) and we have
∫
√
√
∫ 3 1 − sin2 (θ)3 cos(θ)
x 2
3 1 − (3)
dx =
dθ
x
3 sin(θ)
∫
3 cos2 (θ)
=
dθ
sin(θ)
∫
1 − sin2 (θ)
=
dθ
sin(θ)
∫
∫
= csc(θ)dθ − sin(θ)dθ
= − ln | csc(θ) + cot(θ)| + cos(θ) + C
(c)
∫
7
√
dx
2x2 − 3
Similarly this looks like sec2 (θ) − 1 the most so we make it look like this as follows:
√
∫
7
√
√ ( √2x )2
√
3
−1
3
if we let
√
√2x
3
= sec(θ) and dx =
√
√3
2
∫ 7 √3 sec(θ) tan(θ)dθ
√2 √
dx =
3 sec2 (θ) − 1
sec(θ) tan(θ)dθ.
∫
=
7
√ sec(θ)dθ
2
7
= √ ln | sec(θ) + tan(θ)| + C
2
√
2x √2x2 − 3 7
√
= √ ln √ +
+C
2
3
3 4