Math 115 Formulas Sheet and Integration Techniques (Basic)
MATH 115 (S1)1
The following three groups of formulas are the most basic and frequently used formulas in Math 115,
please always keep in mind, especially the underline formulas.
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Basic Trigonometric Identities
sin x
(1) tan x = cos
x,
x
(2) cot x = cos
sin x ,
(3) sec x = cos1 x ,
(4) csc x = sin1 x ,
(5) sin(−x) = − sin x,
(6) cos(−x) = cos x,
(7) sin 2x = 2 sin x cos x,
(8) sin2 x + cos2 x = 1,
(9) sec2 x = tan2 x + 1, (10) sin2 x =
(11) cos2 x =
1−cos 2x
,
2
1+cos 2x
,
2
(12) cos 2x = 2 cos2 x − 1 = 1 − 2 sin2 x = cos2 x − sin2 x
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Basic Derivative Formulas
3
= (k)′ = 0, where k is a number.
(2)
d
(xn )
dx
= (xn )′ = nxn−1 , where n is a number.
= (e x )′ = e x .
(4)
d
(a x )
dx
= (a x )′ = a x ln a, where a > 0, a , 1.
d
(sin
dx
x) = (sin x)′ = cos x.
(6)
d
(cos
dx
x) = (cos x)′ = − sin x.
(7)
d
(tan
dx
x) = (tan x)′ = sec2 x.
(8)
d
(cot
dx
x) = (cot x)′ = − csc2 x.
(9)
d
(sec
dx
x) = (sec x)′ = sec x tan x.
(10)
d
(csc
dx
(1)
d
(k)
dx
(3)
d
(e x )
dx
(5)
x) = (csc x)′ = − csc x cot x.
(11)
d
(sin−1
dx
x) = (sin−1 x)′ =
√1 .
1−x2
(12)
d
(cos−1
dx
x) = (cos−1 x)′ = − √ 1 2 .
(13)
d
(tan−1
dx
x) = (tan−1 x)′ =
1
.
1+x2
(14)
d
(cot−1
dx
1
x) = (cot−1 x)′ = − 1+x
2.
(15)
d
(ln
dx
(16)
d
(loga
dx
x) = (ln x)′ = 1x .
1−x
x) = (loga x)′ =
1
,
x ln a
where a > 0, a , 1.
Basic Indefinite Integral Formulas
(1)
(3)
(5)
(7)
∫
∫
∫
∫
xn dx =
xn+1
n+1
∫
(2)
+ C, where n , −1.
(4)
∫
sin x dx = − cos x + C.
(6)
sec2 x dx = tan x + C.
(8)
∫
∫
kdx = kx + C, where k is a number.
1
x
dx = ln |x | + C.
cos x dx = sin x + C.
csc2 x dx = − cot x + C.
∫
sec x tan x dx = sec x + C.
(10) csc x cot x dx = − csc x + C.
∫
∫
x
(11) e x dx = e x + C.
(12) a x dx = lna a + C, where a > 0, a , 1.
∫
∫ 1
−1
(13) √ 1 2 dx = sin−1 x + C = − cos−1 x + C.
(14) 1+x
x + C = − cot−1 x + C.
2 dx = tan
1−x
∫
∫
(15) sec x dx = ln | sec x + tan x | + C.
(16) csc x dx = − ln | csc x + cot x | + C.
(9)
1
∫
1dx = x + C.
c
⃝April
2010 by Long Yu.
For pdf file, download from:
http://www.math.ualberta.ca/˜longyu/
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4
Main Integration Techniques
4.1
Substitution Rule
(1) First method to use Substitution Rule
∫
∫
′
f [g(x)]g (x)dx =
f (u)du
= F(u) + C
= F[g(x)] + C.
Let u = g(x), then du = g′ (x)dx, replace g′ (x)dx by du.
Integrate the function f (u) with respect to u.
Replace u by g(x) back.
(2) Second method to use Substitution Rule (Make up differential)
∫
∫
′
f [g(x)]g (x)dx =
f [g(x)]d[g(x)]
Make up differential: g′ (x)dx = d[g(x)].
∫
=
f (u)du
Make Substitution: Let u = g(x), then replace all g(x) by u.
= F(u) + C
= F[g(x)] + C.
4.2
Integrate the function f (u) with respect to u.
Replace u by g(x) back.
Integration by Parts
∫
′
∫
f (x)g (x)dx =
f (x)d[g(x)]
∫
= f (x)g(x) − g(x)d[ f (x)]
∫
= f (x)g(x) − g(x) f ′ (x)dx.
Make up differential: g′ (x)dx = d[g(x)].
Use Integration by Parts Formula.
Calculate differential: d[f(x)]=f’(x)dx.
∫
Note:∫ if you use “Integration by Part” correctly, g(x) f ′ (x)dx will be an easier integral than the original
integral f (x)g′ (x)dx, then we can use other methods evaluate it, sometimes we may need to use integration
by parts more than once.
Tips for Integration by Parts:
(1) If the integrand is product of two functions, only one factor can be made up differential, then use that
factor make up differential.
(2) Two methods to make up differential, don’t use power function.
(3) If the integrand is just one factor, don’t make up differential. And if other methods don’t work for the
integral, then
∫
∫
f (x)dx = x f (x) −
x d[ f (x)]
∫
= x f (x) −
Don’t make up differential!
Treat dx as d[g(x)], use “integration by parts” directly.
x f ′ (x)dx
Calculate differential.
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(4) “ILPET”, prefer the latter.
I: Inverse trigonometric function;
L: Logarithmic function;
P: Power function;
E: Exponential function;
T: Trigonometric function.
Explain: If the integrand is the product of two types of functions from “ILPET”, when we make up
differential, prefer to choose the latter function in the above order to make up differential.
4.3
Trigonometric Substitution
4.4
Rational Function Integration
√
(1) a2 − x2 or a2 − x2 , use trigonometric substitution x = a sin θ, where θ ∈ [− π2 , π2 ], or (− π2 , π2 ) (some√
times a2 − x2 may be on the denominator).
√
(2) a2 + x2 or a2 + x2 , use trigonometric substitution x = a tan θ, where θ ∈ (− π2 , π2 ).
√
(3) x2 − a2 or x2 − a2 , use trigonometric substitution x = a sec θ, where θ ∈ [0, π2 ), or (0, π2 ) (sometimes
√
x2 − a2 may be on the denominator).
f (x)
is a proper rational function, that is to say, all f (x) and g(x) are polynomials, and the degree of f (x)
If g(x)
is less than the degree of g(x), i.e. ∂( f (x)) < ∂(g(x)). In the final exam, we only require the following three
cases.
The following are some basic partial fractions patterns (according to the factor of the denominator).
(1) If the denominator can be factored into different linear factors, for example, g(x) = (x+a)(x+b)(x+c),
then
f (x)
A
B
C
=
+
+
.
g(x) x + a x + b x + c
(2) If the denominator can be factored into repeated linear factors, for example, g(x) = (x + a)(x + b)k ,
where (k ≥ 2), then
f (x)
A
B1
B2
Bk
=
+
+
+
·
·
·
+
.
g(x) x + a x + b (x + b)2
(x + b)k
Remember: In the above partial fractions, we have k terms from (x + b)k .
(3) If the denominator can be factored into a linear factor and an irreducible quadratic term: g(x) =
(x + a)(x2 + bx + c), where x2 + bx + c can not be factored into linear factors (the discriminant of x2 + bx + c
is ∆ = b2 − 4c < 0), then
A
Bx + C
f (x)
=
+ 2
.
g(x) x + a x + bx + c
1
1
c
⃝April
2010 by Long Yu.
For pdf file, download from:
http://www.math.ualberta.ca/˜longyu/
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