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Integration by Subsitution II |
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1. Theory
Consider an integral of the form:
Letting u = ƒ(x), then dudx = ƒ′(x), and this gives: du = ƒ′(x) dx.
So:
| ∫ ƒ′(x)ƒ(x) dx | = | ∫ 1ƒ(x)⋅ƒ′(x) dx |
| = | ∫ duu | |
| = | ln u + C |
The general result is:
| ∫ ƒ′(x)ƒ(x)dx = ln ƒ(x) + C |
This is a useful generalisation of the standard integral:
that applies straightforwardly when the top line of what we are integrating is a multiple of the derivative of the bottom line.
2. Exercises
Perform the following integrations.
Click on the questions to reveal their solution
Exercise 1:
Solution:
∫ 10x + 35x2 + 3x − 1 dx is of the form ∫ ƒ′(x)ƒ(x) = ln ƒ(x) + C
Let u = 5x2 + 3x − 1 then dudx = 10x + 3 and du = (10x + 3) dx
| ∴ ∫ 10x + 35x2 + 3x − 1 dx | = | ∫ duu |
| = | ln u + C | |
| = | ln 5x2 + 3x − 1 + C |
Exercise 2:
Solution:
∫ 3x2x3−6 dx is of the form ∫ ƒ′(x)ƒ(x) = ln ƒ(x) + C
Let u = x3 − 6 then dudx = 3x2 and du = (3x2) dx
| ∴ ∫ 3x2x3 − 6 dx | = | ∫ 1u du |
| = | ln u + C | |
| = | ln x3 − 6 + C |
Exercise 3:
Solution:
∫ 4x2x3−1 dx is of the slightly more general form ∫ k ƒ′(x)ƒ(x) = k ∫ ƒ′(x)ƒ(x) = kln ƒ(x) + C
where k is a constant, i.e. the top line is a multiple of the derivative of the bottom line.
Let u = x3 − 1 then dudx = 3x2 and du3 = (x2) dx
| ∴ 4 ∫ x2x3−1 dx | = | 4 ∫ 1u du3 |
| = | 43ln u + C | |
| = | 43ln x3−1 + C |
Exercise 4:
Solution:
∫ 4x2x3−1 dx is of the form ∫ ƒ′(x)ƒ(x) = ln ƒ(x) + C
Let u = x2 − x − 7 then dudx = 2x − 1 and du = (2x − 1) dx
| ∴ ∫ 2x − 1x2 − x − 7 dx | = | ∫ duu |
| = | ln u + C | |
| = | ln x2 − x − 7 + C |
Exercise 5:
Solution:
∫ x − 1x2 − 2x + 7 dx is of the slightly more general form ∫ k ƒ′(x)ƒ(x) = k ∫ ƒ′(x)ƒ(x) = k ln ƒ(x) + C
where k is a constant, i.e. the top line is a multiple of the derivative of the bottom line.
Let u = x2 − 2x + 7, then dudx = 2x − 2 and du = (x2) dx
so that du2 = (x − 1) dx
| ∴ 4 ∫ x − 1x2 − 2x + 7 dx | = | ∫ 12 duu |
| = | 12 ln u + C | |
| = | 12 ln x2 − 2x + 7 + C |
Exercise 6:
Solution:
Let u = x2 − 4 then dudx = 2x and du2 = (x) dx
| ∴ ∫ 8xx2 − 4 dx | = | 8 ∫ 1u du2 |
| = | 4 ln u + C | |
| = | 4 ln x2 − 4 + C |
Exercise 7:
Solution:
Let u = x3 + 2 then dudx = 3x2 and du3 = (x2) dx
| ∴ ∫ x2x3 + 2 dx | = | ∫ 1u du3 |
| = | 13 ln u + C | |
| = | 13 ln x3 + 2 + C |
Exercise 8:
Solution:
Let u = cos (x) then dudx = − sin (x) and du(−1) = sin (x) dx
| ∴ ∫ sin (x)cos (x) dx | = | ∫ 1u du(−1) |
| = | − ∫ duu | |
| = | − ln u + C | |
| = | − ln cos (x) + C |
Exercise 9:
Solution:
Let u = sin (x) then dudx = cos (x) and du= cos (x) dx
| ∴ ∫ cos (x)sin (x) dx | = | ∫ 1u du |
| = | ln u + C | |
| = | ln sin (x) + C |
Exercise 10:
Solution:
Let u = 5 + sin (x) then dudx = cos (x) and du= cos (x) dx
| ∴ ∫ cos (x)5 + sin (x) dx | = | ∫ 1u du |
| = | ln u + C | |
| = | ln 5 + sin (x) + C |
Exercise 11:
Solution:
Let u = 1 − cosh (x) then dudx = −sinh (x) and −du= sinh (x) dx
| ∴ ∫ sinh (x)1 − cosh (x) dx | = | ∫ 1u (−du) |
| = | − ln u + C | |
| = | − ln 1 − cosh (x) + C |
3. Standard Integrals
| ƒ(x) | ∫ƒ(x) dx | ƒ(x) | ∫ƒ(x) dx | |
|---|---|---|---|---|
| xn | xn+1n+1 (n ≠ −1) | [g(x)]n g'(x) | [g(x)]n+1n+1 (n ≠ −1) | |
| 1x | ln x | g'(x)g(x) | ln g(x) | |
| ex | ex | ax | axln a (a > 0) | |
| sin x | −cos x | sinh x | cosh x | |
| cos x | sin x | cosh x | sinh x | |
| tan x | − ln cosx | tanh x | ln cosh x | |
| cosec x | ln tan x2 | cosech x | ln tanh x2 | |
| sec x | ln sec x + tan x | sec x | 2 tan−1ex | |
| sec2 x | tan x | sec2 x | tanh x | |
| cot x | ln sin x | cot x | ln sinh x | |
| sin2 x | x2 − sin 2x4 | sinh2 x | sinh 2x4 − x2 | |
| cos2 x | x2 + sin 2x4 | cosh2 x | sinh 2x4 + x2 | |
| 1a2 + x2 | 1a tan −1 xa (a > 0) | √a2 + x2 | a22[ sinh−1(xa) + x √a2−x2a2] | |
| 1a2 − x2 | 12a ln a + xa − x (0 < x < a) | √a2−x2 | a22[ sin−1(xa) + x √a2−x2a2] | |
| 1x2 − a2 | 12a ln x − ax + a (x > a > 0) | √x2 − a2 | a22[−cosh−1(xa) + x √x2−a2a2] | |
| 1 √ a2 + x2 | ln x + √a2 + x2a (a > 0) | |||
| 1 √ a2 − x2 | sin−1 xa (−a < x < a) | 1 √ x2 − a2 | ln x + √x2 − a2a (x > a > 0) |