Integration Methods (6 pages; 05/09/2013)
(A) Substitution
(A1) ‘By inspection'
Look for 𝑓 ′ (𝑥) in ∫ 𝑓 ′ (𝑥)𝑔(𝑓(𝑥))𝑑𝑥
where ∫ 𝑔(𝑢)𝑑𝑢 can be found
(as 𝑓 ′ (𝑥) will always be in the numerator),
and make the substitution u = f(x).
sin 𝑥
Example: ∫ tan 𝑥 𝑑𝑥 = ∫
𝑑𝑥
cos 𝑥
Here f(x) = cos x and g(u) = −
1
𝑢
1
(A2) Example: ∫ 𝑥(1 + 𝑥)2 𝑑𝑥
Let u = 1+ x
1
[since (u – 1)𝑢2 can be expanded]
(A3) 'speculative'
𝑒𝑥
Example ∫ 2𝑥 𝑑𝑥
𝑒 +1
Let u=𝑒 𝑥 , to give du=𝑒 𝑥 𝑑𝑥 and ∫ = ∫
1
𝑢2 +1
𝑑𝑢 - see (D2)
𝑥
2𝑡
(A4) Substitution of t = tan( ), so that tan x =
2
1−𝑡 2
Referring to the right-angled triangle below,
the hypotenuse = √(1 − 𝑡 2 )2 + 4𝑡 2
1
= √1 + 2𝑡 2 + 𝑡 4 = 1+𝑡 2 (conveniently)
𝑑𝑡
𝑥
1
𝑑𝑥
= 𝑠𝑒𝑐 2 ( ) . , so that
=
𝑑𝑥
2
2
𝑑𝑡
1+𝑡 2
2
𝑥
𝑠𝑒𝑐 2 (2)
=
2
2
1+𝑡 2
1
Example: ∫ 𝑠𝑒𝑐𝑥 𝑑𝑥 = ∫ 2 .
dt = 2∫ 2 𝑑𝑡
1−𝑡 1+𝑡 2
1−𝑡
=∫
1
1−𝑡
1+𝑡 2
= ln|
1
+
1−𝑡 2
1+𝑡
+
2𝑡
1−𝑡 2
1+𝑡
𝑑𝑡 = - ln|1-t| + ln|1+t| = ln|
1−𝑡
| = ln⁡|
1+2𝑡+𝑡 2
1−𝑡 2
|
| = ln|secx + tanx|
(this substitution can sometimes be used to solve trig. equations as well).
(B) Parts
(B1) Example: ∫ ln 𝑥 𝑑x
write as ∫ 1 . ln 𝑥 𝑑x (and integrate the “1”)
1
(B2) Example: I = ∫ . 𝑙𝑛𝑥𝑑𝑥
𝑥
1
= 𝑙𝑛𝑥. 𝑙𝑛𝑥 − ∫ 𝑙𝑛𝑥. 𝑑𝑥 = (𝑙𝑛𝑥)2 − 𝐼
𝑥
1
Hence 𝐼 = (𝑙𝑛𝑥)2 + 𝑐
2
2
(C) Reduction formulae
Integration by Parts can sometimes enable a recurrence relation to be set
up.
1
Example: 𝐼𝑛 = ∫0 𝑥 𝑛 𝑒 −𝑥 𝑑𝑥
Integrating by Parts (differentiating 𝑥 𝑛 ) leads to 𝐼𝑛 = 𝑛𝐼𝑛−1 - 𝑒 −1
𝜋
Example: 𝐼𝑛 = ∫0 𝑐𝑜𝑠 𝑛 𝑥 𝑑𝑥 (n even)
Integrating by Parts (writing as 𝑐𝑜𝑠𝑥 . 𝑐𝑜𝑠 𝑛−1 𝑥 and differentiating
𝑐𝑜𝑠 𝑛−1 𝑥)
leads to 𝐼𝑛 =
𝑛−1
𝑛
𝐼𝑛−2
1 𝑥𝑛
Example: 𝐼𝑛 = ∫0
√1−𝑥
𝑑𝑥
Integrating by Parts (differentiating 𝑥 𝑛 ) leads to 𝐼𝑛 =
2𝑛𝐼𝑛−1
2𝑛+1
Notes
(i) Parts may need to be applied twice
(ii) The formula for 𝐼𝑛 could involve 𝐼𝑛−1 and 𝐼𝑛−2
(iii) In order to involve 𝐼𝑛−1 or 𝐼𝑛−2 , it may be necessary to rearrange the
integrand
(iv) Sometimes the integrand can be rearranged to give the Reduction
formula, without performing integration by Parts
eg ∫ 𝑡𝑎𝑛𝑛 𝑥𝑑𝑥 = ∫ 𝑡𝑎𝑛𝑛−2 𝑥(𝑠𝑒𝑐 2 𝑥 − 1)𝑑𝑥 𝑒𝑡𝑐 (since 𝑠𝑒𝑐 2 𝑥 is the derivative
of tanx)
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(D) Rearranging the integrand
(D1) Example: ∫
1+𝑥
𝑥−1
𝑑𝑥 = ∫
𝑥−1
𝑥−1
𝑑𝑥 + ∫
2
𝑥−1
𝑑𝑥 etc
(D2) Example: ∫ 𝑡𝑎𝑛2 𝑥 dx = ∫ 𝑠𝑒𝑐 2 𝑥 − 1 𝑑𝑥
= tan x - x + c
𝑠𝑖𝑛ℎ𝑥
(D3) Example: ∫ 𝑠𝑒𝑐ℎ𝑥𝑡𝑎𝑛ℎ𝑥 𝑑𝑥 = ∫
𝑑𝑥
𝑐𝑜𝑠ℎ 2 𝑥
(D4) Partial Fractions
1
1
Example: ∫ 2 2 𝑑𝑥 = ∫
𝑑𝑥 etc
𝑥 −𝑎
(𝑥−𝑎)(𝑥+𝑎)
(D5) Completing the square
1
1
Example: ∫ 2
𝑑𝑥 = ∫ (𝑥+3)2
𝑑𝑥 ; then see (E2)(b)
𝑥 +6𝑥+13
+4
(D6) Using a definition
Example: ∫ 𝑠𝑒𝑐ℎ𝑥𝑑𝑥 = ∫
𝑒𝑥
2
𝑑𝑥 = 2 ∫ 2𝑥 𝑑𝑥 (then subst. u=𝑒 𝑥 )
𝑒 𝑥 +𝑒 −𝑥
𝑒 +1
(E) Standard integrals
(E1) Using standard derivatives
(a)
𝑑
𝑑𝑥
(tan x) = 𝑠𝑒𝑐 2 x , so that ∫ 𝑠𝑒𝑐 2 𝑥 dx = tan x + c
𝑑
(b) To find ∫ 𝑐𝑜𝑠𝑒𝑐ℎ2 𝑥 𝑑𝑥, note that (tanh x) = 𝑠𝑒𝑐ℎ2 x, and establish
𝑑𝑥
that
𝑑
𝑑𝑥
(coth x) = -co𝑠𝑒𝑐ℎ2 x
[This is an example of the general approach of looking for something
related to the integrand in question.]
4
(E2) Inverse trigonometric functions
1
𝑥
(a) ∫ 2 2 𝑑𝑥 = 𝑠𝑖𝑛−1 ( ) + c (x<|a|)
𝑎
√𝑎 −𝑥
Proof: Substitute x=asinx, or establish derivative of 𝑠𝑖𝑛−1 𝑥
(𝑏) ∫
1
𝑎2 +𝑥 2
𝑑𝑥 =
1
𝑥
𝑡𝑎𝑛−1 ( ) + 𝑐
𝑎
𝑎
1
𝑥
Suppose that ∫ 2 2 𝑑𝑥 = 𝐴 𝑎𝑟𝑐𝑡𝑎𝑛 ( ) + 𝐶
𝑎 +𝑥
𝑎
Then
𝑑
𝑑𝑥
𝑥
𝐴 𝑎𝑟𝑐𝑡𝑎𝑛 ( ) = A .
𝑎
Hence A =
1
𝑥
1+( )2
𝑎
.
1
𝑎
=
𝐴𝑎
𝑎2 +𝑥 2
1
𝑎
(E3) Inverse Hyperbolic functions
1
𝑥
(a) ∫ 2 2 𝑑𝑥 = 𝑠𝑖𝑛ℎ−1 ( ) + c
𝑎
√𝑎 +𝑥
(b) ∫
1
√𝑥 2 −𝑎2
𝑥
𝑑𝑥 = 𝑐𝑜𝑠ℎ−1 ( ) + c (x>a)
𝑎
Note: As for ∫
1
√𝑎2 −𝑥 2
𝑥
𝑑𝑥 = 𝑠𝑖𝑛−1 ( ), A=1 (referring to (E2))
𝑎
(F) Sines & Cosines
(F1) Powers of sines & cosines
(a)Odd powers: Example: ∫ 𝑠𝑖𝑛7 𝑥 𝑑𝑥
𝑠𝑖𝑛7 𝑥 = 𝑠𝑖𝑛𝑥 . (1 − 𝑐𝑜𝑠 2 𝑥)3 , which can be expanded to give a series of
integrals of the form 𝑠𝑖𝑛𝑥 . 𝑐𝑜𝑠 𝑟 𝑥 [see (A1)]
and similarly for odd powers of cosx
(b) Even powers: See (C) for use of Reduction Formulae
5
(F2) ∫ sin(mx)cos(nx)dx
As sin(mx±nx) = sin(mx)cos(nx) ± cos(mx)sin(nx),
sin(mx)cos(nx) = ½ {sin(mx+nx) + sin(mx-nx)}
(F3) Convert sine to cosine using sin x = cos (
π
2
− x)
Example: ∫ √1 + sin2x dx
π
π
π
2
4
4
1+sin2x = 1+cos( − 2x) = 1+cos2( − x) = 2cos 2 ( − x) etc
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