# 1. Theory

Consider an integral of the form:

[ƒ(x)]nƒ′(x) dx

Letting u = ƒ(x) gives du/dx = ƒ′(x), and du = ƒ′(x) dx.

 ∫  [ƒ(x)]nƒ′(x) dx = ∫  un du = un + 1/n + 1 + C = [ƒ(x)]n + 1/n + 1 + C

For example, when n = 1:

ƒ(x)ƒ′(x) dx =     udu = u2/2 + C = [ƒ(x)]2/2 + C

# 2. Exercises

Perform the following integrations.

Click on the questions to reveal their solution

Exercise 1:

sin (x) cos (x) dx

Solution:

sin (x) cos (x) dx is of the form     ƒ(x)ƒ′(x) dx = 1/2 [ƒ(x)]2 + C

To see this, set u = sin (x), and find du/dx = cos (x) and du = cos (x) dx

 ∴    ∫  sin (x)cos (x) dx = ∫  u du = 1/2 u2 + C = 1/2 sin2(x) + C

Exercise 2:

sinh (x) cosh (x) dx

Solution:

sinh (x) cosh (x) dx is of the form     ƒ(x)ƒ′(x) dx = 1/2 [ƒ(x)]2 + C

To see this, set u = sinh (x), and find du/dx = cosh (x) and du = cosh (x) dx

 ∴    ∫  sinh (x)cosh (x) dx = ∫  u du = 1/2 u2 + C = 1/2 sinh2(x) + C

Exercise 3:

tan (x) sec2 (x) dx

Solution:

tan (x) sec2(x) dx is of the form     ƒ(x)ƒ′(x) dx = 1/2 [ƒ(x)]2 + C

To see this, set u = tan (x), and find du/dx = sec2(x) and du = sec2(x) dx

 ∴    ∫ tan (x) sec2 (x) dx = ∫  u du = 1/2 u2 + C = 1/2 tan2 (x) + C

Exercise 4:

sin (2x) cos (2x) dx

Solution:

sin (2x) cos (2x) dx is close to the form     ƒ(x)ƒ′(x) dx = 1/2 [ƒ(x)]2 + C

To see this, set u = sin (2x), and find du/dx = 2 cos (2x) and du2 = cos (2x) dx

 ∴    ∫  sin (2x) cos (2x) dx = ∫  udu2 = 1/2   ∫ u du = 1/2⋅1/2u2 + C = 1/4 sin2 (2x) + C

Exercise 5:

sinh (3x) cosh (3x) dx

Solution:

sinh (3x) cosh (3x) dx is close to the form     ƒ(x)ƒ′(x) dx = 1/2 [ƒ(x)]2 + C

To see this, set u = sinh (3x), and find du/dx = 3 cosh (3x) and du3 = cosh (3x) dx

 ∴    ∫  sinh (3x) cosh (3x) dx = ∫  u du3 = 1/3   ∫  u du = 1/3⋅1/2 u2 + C = 1/6 sinh2 (3x) + C

Exercise 6:

1/xln(x) dx, x > 0

Solution:

1/xln(x) dx is of the form     ƒ(x)ƒ′(x) dx = 1/2 [ƒ(x)]2 + C

To see this, set u = ln (x), and find du/dx = 1/x and du = 1/xdx

 ∴   ∫  1/xln (x) dx = ∫ u du = 1/2 u2 + C = 1/2 ln2 (x) + C

Exercise 7:

sin4 (x) cos (x) dx

Solution:

sin4 (x) cos (x) dx is of the form     [ƒ(x)]nƒ′(x) dx = [ƒ(x)]n + 1/n + 1 + C

To see this, set u = sin (x), and find du/dx = cos (x) and du = cos (x) dx

 ∴    ∫ sin4 (x) cos (x) dx = ∫  u4 du = u5/5 + C = sin5 (x)/5 + C

Exercise 8:

sinh3 (x) cosh (x) dx

Solution:

sinh3(x) cosh (x) dx is of the form     [ƒ(x)]nƒ′(x) dx = [ƒ(x)]n+1/n+1 + C

To see this, set u = sinh (x), and find du/dx = cosh (x) and du = cosh (x) dx

 ∴    ∫ sinh3 (x) cosh (x) dx = ∫ u3 du = u4/4 + C = sinh4 (x)/4 + C

Exercise 9:

cos3 (x) sin (x) dx

Solution:

cos3 (x) sin (x) dx is close to the form     [ƒ(x)]nƒ′(x) dx = [ƒ(x)]n + 1/n + 1 + C

To see this, set u = cos (x), and find du/dx = − sin (x) and du = − sin (x) dx

 ∴    ∫  cos3 (x) sin (x) dx = ∫  u3⋅(−du) = −   ∫  u3 du = − u4/4 + C = − cos4 (x)/4 + C

Exercise 10:

2x/(x2 − 4)2dx

Solution:

2x/(x2 − 4)2 dx is of the form     [ƒ(x)]nƒ′(x) dx = [ƒ(x)]n + 1/n + 1 + C

To see this, set u = x2 − 4, and find du/dx = 2x and du = 2xdx

 ∴    ∫  2x/(x2 − 4)2 dx = ∫  1/u2 = −   ∫  u − 2 du = −u−1 = − 1/u + C = − 1/x2 − 4 + C

# 3. Standard Integrals

ƒ(x) ∫ƒ(x) dx ƒ(x) ∫ƒ(x) dx xn xn+1/n+1 (n ≠ −1) [g(x)]n g'(x) [g(x)]n+1/n+1 (n ≠ −1) 1/x ln x g'(x)/g(x) ln g(x) ex ex ax ax/ln 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  x/2 cosech x ln tanh  x/2 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 x/2 − sin 2x/4 sinh2 x sinh 2x/4 − x/2 cos2 x x/2 + sin 2x/4 cosh2 x sinh 2x/4 + x/2 1/a2 + x2 1/a tan −1  x/a   (a > 0) √a2 + x2 a2/2[ sinh−1(x/a) + x √a2−x2/a2] 1/a2 − x2 1/2a ln  a + x/a − x    (0 < x < a) √a2−x2 a2/2[ sin−1(x/a) + x √a2−x2/a2] 1/x2 − a2 1/2a ln  x − a/x + a   (x > a > 0) √x2 − a2 a2/2[−cosh−1(x/a) + x √x2−a2/a2] 1/ √ a2 + x2 ln  x + √a2 + x2/a   (a > 0) 1/ √ a2 − x2 sin−1  x/a     (−a < x < a) 1/ √ x2 − a2 ln  x + √x2 − a2/a   (x > a > 0)