Download Integration Formulas and Techniques for Math 305 - Prof. Jianhong Xu and more Study notes Mathematics in PDF only on Docsity!
Math 305 Methods of Integration
The following are a list of integration formulas that you should know. Even when you go through the different methods, you reduce the integrals to one of these:
xndx = xn+ n + 1
sec^2 xdx = tan x + c
dx x
= ln |x| + c
csc^2 xdx = − cot x + c
∫ exdx = ex^ + c
sec x tan xdx = sec x + c
∫ sin xdx = − cos x + c
csc x cot xdx = − csc x + c
∫ cos xdx = sin x + c
The following are nice to know (make your life easier) integrals:
tan xdx = − ln | cos x| + c
csc xdx = − ln | csc x + cot x| + c
cot xdx = ln | sin x| + c
sec xdx = ln | sec x + tan x| + c
If a is constant, a 6 = 0 :
eaxdx =
a
eax^ + c
cos axdx =
a
sin ax + c
sin axdx = −
a
cos ax + c
Hint I: If you think you have an antiderivative, just differentiate and see if you end up with the integrand!
Hint II: The line of attack you should take when trying to integrate should be to try these methods in this order. You will always need to have paper and pencil, the correct method doesn’t usually bounce off the page and hit you in the face!
Method 1. Straight forward integration. The integral can be reduced to one of the first set of integrals.
Example A.
(x^2 + 1)(x − 3)dx =
(x^3 − 3 x^2 + x − 3)dx =
x^4 4
− x^3 +
x^2 2
− 3 x + c
Example B.
1 − sin^2 x 1 + sin x
dx =
(1 − sin x)(1 + sin x) 1 + sin x
dx =
(1 − sin x)dx = x + cos x + c
Problems:
ex(1 − e−x^ sec^2 x)dx
x^3 − 2 x + 4 x
dx
cos x tan xdx
Method 2. Substitution. A simple substitution reduces the integral to one of the first set of integrals. In fact, the second set came about by using substitution.
Example C.
tan xdx =
sin x cos x
dx. Let w = cos x then dw = − sin xdx so that the integral becomes ∫
−
dw w
= − ln |w| + c = − ln | cos x| + c.
Example D.
xex
2 dx. Let w = x^2 then dw = 2xdx so that the integral becomes
ew^
dw 2
ew^ + c = 1 2
ex
2
Hint III: Things to look for: if the integrand involves
ef^ (x), trig (f(x)),
f(x)
, (f(x))n
Let w = f(x). This is not an exclusive list!
Hint IV: When substitution is complete, make sure no x’s appear in the integrand.
Hint V: When choosing u and dv make sure dv is something that can be integrated. Also, the whole integrand should be taken up with u and dv.
Hint VI: Method should be used when integrand involves (poly)trig eax(sin bx) eax^ cos bx sec(2n+1)^ x (poly) ln x poly(inverse trig fact) csc(2n+1)^ x
This is not an exclusive list!
Problems:
ln xdx. Let u = ln x, dv = dx.
(x^2 + 2x − 1) cos 3xdx. Let u = x^2 + 2x − 1 , dv = cos 3xdx. (Need to use integration by parts twice.)
(x + 3)e^2 xdx. Let u = x + 3, dv = e^2 xdx.
x^2 ln xdx
tan−^1 xdx
ex^ sin xdx
Method 4. Trigonometric integrals. Integrands only involve trigonometric functions (not inverse trig functions!). Remember that certain trig functions “go together.”
sin θ and cos θ tan θ and sec θ cot θ and cot θ
If you have mixed trig functions convert everything to sin θ and cos θ. Another helpful identity is sin^2 θ + cos^2 θ = 1. From here you can derive tan^2 θ + 1 = sec^2 θ and 1 + cot^2 θ = csc^2 θ. You should also have the half angle formulas in your repertoire.
sin^2 θ =
(1 − cos 2θ)
cos^2 θ =
(1 + cos 2θ)
Most trigonometric integrations involve substitution. The only exceptions are when the integrand is sec^2 n+1^ θ or csc^2 n+1^ θ (see integration by parts) or sin^2 n^ θ cos^2 m^ θ which involves the half angle formula (sometimes repeated several times).
Example G.
sin θ cos^2 θdθ. Let w = cos θ; dw = − sin θdθ ∫ sin θ cos^2 θdθ =
−w^2 dw = −
w^3 3
cos^3 θ 3
Example H.
tan^2 θdθ =
(sec^2 θ − 1)dθ = tan θ − θ + c
Example I.
sin^2 θ cos^3 θdθ Let w = sin θ then dw = cos θdθ.
∫ sin^2 θ cos^2 θ sin︸ ︷︷ θdθ ︸ dw
w^2 (1 − sin^2 θ)dw =
w^2 (1 − w^2 )dw =
w^3 3
w^5 5
sin^3 θ 3
sin^5 θ 5
Example J.
sin^2 θdθ =
(1 − cos 2θ)dθ =
θ −
sin 2θ + c
Hint VII: As you can see there’s not a whole lot of guessing to do with substitution. For instance in Example I above if you had let w = cos θ then dw = − sin θdθ and you have an “extra” sin θ in the integrand. Of course it would be silly to guess w = tan θ or sec θ or another trig function. Use common sense and a paper and pencil!
Problems:
tan θ sec^3 θdθ 19.
sin θ cos θdθ
cos^2 θdθ 20.
tan^3 2 θ sec^2 2 θdθ
cot θ csc^3 θdθ 21.
sin^3 θ cos^4 θdθ
Note that integrands involving different arguments were not covered. i.e.,
cos 2x sin 3xdx. These type
of integrals can be done using integration by parts twice and bringing the integral to the other side.
Method 6. Partial Fractions. For this method the integrand can be written in the form
polynomial polynomial
Steps:
- Is the degree of the polynomial in the denominator greater than the degree of the polynomial in the numerator? If not, divide out.
- Factor denominator.
- Write the fraction as a sum of fractions.
- Solve for constants.
- Integrate.
Example M.
x − 4 x^2 + 5x + 6
dx
- Yes.
- x^2 + 5x + 6 = (x + 2)(x + 3)
x − 4 x^2 + 5x + 6
A
x + 2
B
x + 3
- x − 4 = A(x + 3) + B(x + 2)
Let x = − 3 − 3 − 4 = B(−3 + 2) ⇒ B = 7
Let x = − 2 − 2 − 4 = A(−2 + 3) ⇒ A = − 6
(x − 4) x^2 + 5x + 6
dx =
∫ (^
x + 2
x + 3
dx = −6 ln |x + 2| + 7 ln |x + 3| + c
Example N.
(x^2 + 2x − 4) x^3 + x
dx
Steps:
- Yes.
- x^3 + x = (x + 1)(x^2 − x + 1)
x^2 + 2x − 4 (x + 1)(x^2 − x + 1)
A
x + 1
Bx + C x^2 − x + 1
- x^2 + 2x − 4 = A(x^2 − x + 1) + (Bx + C)(x + 1) Let x = − 1 1 − 2 − 4 = A(3) ⇒ A =
Set coefficients equal
coef of x^2 1 = A + B coef of x 2 = −A + B + C const − 4 = A + C
Using A = −
, C = −
, B =
x^2 + 2x − 4 x^3 + x
dx =
x + 1
dx +
8 x − 7 x^2 − x + 1
dx
to integrate the 2nd integral x^2 − x + 1 =
x −
so that
8 x − 7 1 4
x −
) 2 dx.
Let w = x −
and use trig substitution.
Problems:
(x^2 − 4)dx x^3 + 3x^2 + 2x
x + 3 x^3 + x
dx
x − 3 x^3 + 4x^2 + 4x
dx
x^3 − x
dx
