Integration by Subsitution III |
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Consider an integral of the form:
Letting u = ƒ(x) gives dudx = ƒ′(x), and du = ƒ′(x) dx.
∫ [ƒ(x)]nƒ′(x) dx | = | ∫ un du |
= | un + 1n + 1 + C | |
= | [ƒ(x)]n + 1n + 1 + C |
For example, when n = 1:
Perform the following integrations.
Click on the questions to reveal their solution
Exercise 1:
Solution:
∫ sin (x) cos (x) dx is of the form ∫ ƒ(x)ƒ′(x) dx = 12 [ƒ(x)]2 + C
To see this, set u = sin (x), and find dudx = cos (x) and du = cos (x) dx
∴ ∫ sin (x)cos (x) dx | = | ∫ u du |
= | 12 u2 + C | |
= | 12 sin2(x) + C |
Exercise 2:
Solution:
∫ sinh (x) cosh (x) dx is of the form ∫ ƒ(x)ƒ′(x) dx = 12 [ƒ(x)]2 + C
To see this, set u = sinh (x), and find dudx = cosh (x) and du = cosh (x) dx
∴ ∫ sinh (x)cosh (x) dx | = | ∫ u du |
= | 12 u2 + C | |
= | 12 sinh2(x) + C |
Exercise 3:
Solution:
∫ tan (x) sec2(x) dx is of the form ∫ ƒ(x)ƒ′(x) dx = 12 [ƒ(x)]2 + C
To see this, set u = tan (x), and find dudx = sec2(x) and du = sec2(x) dx
∴ ∫ tan (x) sec2 (x) dx | = | ∫ u du |
= | 12 u2 + C | |
= | 12 tan2 (x) + C |
Exercise 4:
Solution:
∫ sin (2x) cos (2x) dx is close to the form ∫ ƒ(x)ƒ′(x) dx = 12 [ƒ(x)]2 + C
To see this, set u = sin (2x), and find dudx = 2 cos (2x) and du2 = cos (2x) dx
∴ ∫ sin (2x) cos (2x) dx | = | ∫ udu2 |
= | 12 ∫ u du | |
= | 12⋅12u2 + C | |
= | 14 sin2 (2x) + C |
Exercise 5:
Solution:
∫ sinh (3x) cosh (3x) dx is close to the form ∫ ƒ(x)ƒ′(x) dx = 12 [ƒ(x)]2 + C
To see this, set u = sinh (3x), and find dudx = 3 cosh (3x) and du3 = cosh (3x) dx
∴ ∫ sinh (3x) cosh (3x) dx | = | ∫ u du3 |
= | 13 ∫ u du | |
= | 13⋅12 u2 + C | |
= | 16 sinh2 (3x) + C |
Exercise 6:
Solution:
∫ 1xln(x) dx is of the form ∫ ƒ(x)ƒ′(x) dx = 12 [ƒ(x)]2 + C
To see this, set u = ln (x), and find dudx = 1x and du = 1x dx
∴ ∫ 1xln (x) dx | = | ∫ u du |
= | 12 u2 + C | |
= | 12 ln2 (x) + C |
Exercise 7:
Solution:
∫ sin4 (x) cos (x) dx is of the form ∫ [ƒ(x)]nƒ′(x) dx = [ƒ(x)]n + 1n + 1 + C
To see this, set u = sin (x), and find dudx = cos (x) and du = cos (x) dx
∴ ∫ sin4 (x) cos (x) dx | = | ∫ u4 du |
= | u55 + C | |
= | sin5 (x)5 + C |
Exercise 8:
Solution:
∫ sinh3(x) cosh (x) dx is of the form ∫ [ƒ(x)]nƒ′(x) dx = [ƒ(x)]n+1n+1 + C
To see this, set u = sinh (x), and find dudx = cosh (x) and du = cosh (x) dx
∴ ∫ sinh3 (x) cosh (x) dx | = | ∫ u3 du |
= | u44 + C | |
= | sinh4 (x)4 + C |
Exercise 9:
Solution:
∫ cos3 (x) sin (x) dx is close to the form ∫ [ƒ(x)]nƒ′(x) dx = [ƒ(x)]n + 1n + 1 + C
To see this, set u = cos (x), and find dudx = − sin (x) and du = − sin (x) dx
∴ ∫ cos3 (x) sin (x) dx | = | ∫ u3⋅(−du) = − ∫ u3 du |
= | − u44 + C | |
= | − cos4 (x)4 + C |
Exercise 10:
Solution:
∫ 2x(x2 − 4)2 dx is of the form ∫ [ƒ(x)]nƒ′(x) dx = [ƒ(x)]n + 1n + 1 + C
To see this, set u = x2 − 4, and find dudx = 2x and du = 2x dx
∴ ∫ 2x(x2 − 4)2 dx | = | ∫ 1u2 = − ∫ u − 2 du |
= | −u−1 = − 1u + C | |
= | − 1x2 − 4 + C |
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) |