NCSSM Online AP Calculus (AB)
Distance Education & Extended Programs
Lesson 4.8 ... Last Anti-­‐Derivative Rules and u-­‐Substitution RE-VISIT THE INTEGRAL OF 1/x
We saw earlier this unit that
1
∫ x dx = ln x + C .
k
If we consider the definite integral
x = 1 to x = k.
1
∫ x dx , you can think of that as the area under the curve
1
y=
1
from
x
Suppose, then, that k were to equal e . Then the area under the curve y =
1
would be ln e which
x
equals 1. Interesting! J
ANTI-DERIVATIVE RULES OF THE REMAINING TRIG FUNCTIONS
At the beginning of the unit, we saw some of the trigonometry anti-derivative rules … the ones that come
directly from the differentiation rules.
Now we look at the anti-derivative rules for the remaining trig functions.
NCSSM Online AP Calculus (AB)
Distance Education & Extended Programs
∫ tan x dx = − ln cos x + C
∫ cot x dx = ln sin x + C
∫ sec x dx = ln sec x + tan x + C
∫ csc x dx = − ln csc x + cot x + C
Notice that the anti-derivatives whose answers begin with the letter “c” are
still negative as we saw with the rules at the start of the unit! J
How and why is it that these anti-derivative rules include the natural logarithm function? See the proofs
of ∫ tan x dx and ∫ sec x dx at the bottom of Page 503 of our textbook. (Don’t worry … they are short
“algebraic” proofs!)
Indefinite and definite integrals that will use these rules are done just as we saw earlier problems
evaluated, including use of u-substitution. Remember that the definite integrals can also be checked by
using the built-in fnInt (MATH-9) function on your graphing calculator!
Now you try some! The worked solutions to these problems can be found at the end of this document.
#1) Evaluate
∫ tan 3x
#2) Evaluate
∫ sin 2x dx
1
π /4
#3) Evaluate
dx
∫
1+ tan 2 x dx
0
π /3
#4) Evaluate
∫
π /6
cot 2x dx
NCSSM Online AP Calculus (AB)
Distance Education & Extended Programs
Problem Solutions:
#1)
∫ tan 3x
dx . Let u = 3x . Then du = 3dx →
1
du = dx . The original problem is re-written, then,
3
1
1
1
tanu du = − ln cosu + C = − ln cos 3x + C . Remember that your answers to these indefinite
∫
3
3
3
integrals can be checked by taking the derivative of your answer and you should obtain the original
integrand back again. If you take the derivative here, you obtain
d ⎛ 1
⎞
⎜⎝ − ln cos 3x + C ⎟⎠
dx
3
1
1
=− ⋅
⋅ − sin 3x ⋅ 3 + 0
3 cos 3x
sin 3x
=
cos 3x
= tan 3x
as
#2)
1
∫ sin 2x dx .
Begin by re-writing this integral as
∫ csc 2x
dx . Let u = 2x . Then du = 2dx →
1
du = dx . The original problem is re-written, then, as
2
1
1
1
cscu du = − ln cscu + cot u + C = − ln csc 2x + cot 2x + C . Again, you could take the derivative of
∫
2
2
2
this answer to make sure you obtain the original integrand back again.
π /4
#3)
∫
1+ tan x dx . Begin by using a Pythagorean Identity to re-write this as
0
can be simplified further to
π /4
∫
0
π /4
2
sec 2 x dx . This
0
π /4
∫
∫
sec x dx . Then
0
π /4
sec x dx = ⎡⎣ ln sec x + tan x ⎤⎦ 0
= ln 2 + 1 − ln 1+ 0 = ln 2 + 1 − 0 = ln
(
)
2 + 1 . You should be able
to do this problem without the use of a calculator, by drawing upon your Unit Circle knowledge!
NCSSM Online AP Calculus (AB)
Distance Education & Extended Programs
π /3
∫
#4)
cot 2x dx . Let u = 2x . Then du = 2dx →
π /6
2π /3
1
du = dx . The original problem is re-written, then,
2
1
3
3⎤ 1
2π /3 1 ⎡
− ln
⎡⎣ ln sinu ⎤⎦π /3 = ⎢ ln
⎥ = (0) = 0 . You should be able to do this
2
2⎣ 2
2 ⎦ 2
π /3
problem without the use of a calculator, by drawing upon your Unit Circle knowledge!
as
1
2
∫
cot u du =