Substitution Method

Example 6.
Evaluate each of the following integrals.
(a) ∫ xe^x² dx
(b) ∫ sin⁴ x cos x dx
(c) ∫ x² ( ^x³⁄₃ + 1 )^² dx
(d) ∫ √ 1 - 5x dx
(e) ∫ 3x √ x² + 5 dx
(f) ∫ cos³ x dx

Solution
(a) ∫ xe^x² dx
Let u = x². Then du = 2x dx,
¹⁄₂ du = x dx.
∫ xe^x² = ∫ e^x² x dx
= ∫ e^u ¹⁄₂ e^u + C
= ¹⁄₂ e^x² + C.

(b) ∫ sin⁴ x cos x dx .
Let u = sin x. Then du = cos x dx .
∫ sin x⁴ cos x dx = ∫ u⁴ du
= ^u⁵⁄₅ + C = ^{sin⁵ x}⁄₅ + C.

(c) ∫ x² ( ^x³⁄₃ + 1 )^² dx
Let u = ^x³⁄₃ + 1. Then du = x² dx.
∫ x² ( ^x³⁄₃ + 1 ) ^³ dx = ∫ u³ du
= ^u⁴⁄₄ + C
= ¹⁄₄ ( ^x³⁄₃ + 1 )^⁴ + C.

(d) ∫ √ 1 - 5x dx
Let u = 1 - 5x. Then du = -5 dx,
- ¹⁄₅ du = dx.
∫ √ 1 - 5x dx = ∫(1 - 5x)^{^¹⁄₂} dx
= ∫ u^{^¹⁄₂} (- ¹⁄₅) du
= - ¹⁄₅ ⋅ ²⁄₃ u^{^³⁄₂} + C
= - ²⁄₁₅ u^{^³⁄₂} + C
= - ²⁄₁₅(1 - 5x)^{^¹⁄
₂} + C.

(e) ∫ 3x √ x² + 5 dx
Let u = x² + 5. Then ¹⁄₂du = x dx
∫ 3x √ x² + 5 dx = 3 ∫ (x² + 5)^{^¹⁄₂} x dx
= 3 ∫ u^{^¹⁄₂} ¹⁄₂ du
= ³⁄₂ ⋅ ²⁄₃ u^{^³⁄₂} + C
= u^{^³⁄₂} + C
= (x² + 5)^{^³⁄₂} + C.

(f) ∫ cos³ x dx

We use cos² x = 1 - sin² x, then
∫ cos³ x dx = ∫ (1 - sin² x) cos x dx.
Let u = sin x, then du = cos x dx
∫p cos³ x dx = ∫ (1 - u² du
= u - ^u³⁄₃ + C
= sin x - ¹⁄₃ sin³ x + c.

2. ∫ ^{g ' (x)}⁄_{g (x)} dx
Let u = g (x). Then du = g ' (x dx.
Thus
∫ ^{g ' (x)}⁄_{g (x)} dx = ∫ ¹⁄_u du = ln |u| + C.

∫ ^{g ' (x)}⁄_{g (x} dx = ln |g (x)| + C.
∫ tan x dx = ∫ ^{sin x}⁄_{cos x} dx = - ln |cos x| + C.
∫ cot x dx = ∫ ^{cos x}⁄_{sin x} dx = ln |sin x| + C.

Example 7.
Evaluate the following integrals.
(a) ∫ ^{2x - 1}⁄_{x² - x - 6} dx
(b) ∫ ^{x² + 2x - 1}⁄_{x² - 1} dx
(c) ∫ ¹⁄_{1 + e^x} dx

Solution
(a) ∫ ^{2x - 1}⁄_{x² - x - 6} dx
Let u = x² - x - 6. Then du = (2x - 1) dx.
∫ ^{2x - 1}⁄_{x² - x - 6} dx = ∫ ¹⁄_u du
= ln |u| + C = ln |x² - x - 6| + C.

(b) ∫ ^{x² + 2x - 1}⁄_{x² - 1} dx
We can rewrite as
∫ ^{x² + 2x - 1}⁄_{x² - 1} dx = ∫ (1 + ^2x⁄_{x² - 1} ) dx
= ∫ dx + ∫ ¹⁄_u du
= x + ln |u| + C
= x + ln |x² - 1| + C.

(c) ∫ ¹⁄_{1 + e^x} dx
We can rewrite by multiplying and dividing by e^-x.
(c) ∫ ¹⁄_{1 + e^x} dx = ∫ ¹⁄_{1 + e^x} ^{e^-x}⁄_e^-x dx
= ∫ ^{e^-x}⁄_{e^-x + 1} dx.
Let u = e^-x + 1. Then du = -e^-x dx.
∫ ¹⁄_{1 + e^x} dx = ∫ ^{e^-x}⁄_{e^-x + 1} dx
= - ∫ ¹⁄_u du
= - ln |u| + C.
= - ln (e^-x + 1) + C.

We now explain some integrals involving the trigonometric functions using method of substitution.

(a) ∫ sec x dx = ln | sec x + tan x + C
Proof
We have
∫ sec x dx = ∫ sec x ^{sec x + tan x}⁄_{sec x + tan x} dx
= ∫ ^{sec² x + sec x tan x}⁄_{sec x + tan x} dx.
Let u = sec x + tan x. Then
du = (sec x tan x + sec² x) dx.
So,
∫ sec x dx = ∫ ¹⁄_u du
= ln |u| + C
= ln |sec x + tan x| + C.

(b) ∫ csc x dx = - ln |csc x + cot x| + C
Proof
We have
∫ csc x dx = ∫ csc x ^{csc x + cot x}⁄_{csc x + cot x} dx
= ∫ ^{csc² x + cot x csc x}⁄_{csc x + cot x} dx.
Let u = csc x + cot x. Then
du = - (cot x csc x + csc² x) dx.
So,
∫ csc x dx = - ∫ ¹⁄_u du = - ln |u| + C
= - ln |csc x + cot x| + C.

Answer to 1.(h)

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Exercise 10.2
1. Integrate the following functions using the given substitutions.
(a) 4x³ √ x⁴ - 1 ; u = x⁴ - 1
Solution

Let u = x⁴ - 1
^du⁄_dx = 4x³
du = 4x³ dx
$$ \int 4x^3 \sqrt{x^4 - 1}\ dx = \int (x^4 - 1)^{\frac{1}{2}}\ 4x^3 \ dx $$ $$= \int u^{\frac{1}{2}} \ du $$ $$= \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}$$ = ^{u^{^³⁄₂}}⁄_³⁄₂ + C
= ²⁄₃ (x⁴ - 1)^{^³⁄₂} + C

(b) cos³ x sin x ; u = cos x
Solution
Let u = cos x
^du⁄_dx = - sin x
du = - sin x dx
= - du = sin x dx
∫ cos³ x sin x dx = ∫ u³ (- du)
= - ∫ u³ du
= - ^u³⁺¹⁄₃₊₁ + C
= - ¹⁄₄ u⁴ + C
= - ¹⁄₄ cos⁴ x + C

(c) ¹⁄_{x ln |x|} ; u = ln |x|
Solution
Let u = ln |x|
^du⁄_dx = ¹⁄_|x|
du = ¹⁄_x dx, x > 0
or du = ¹⁄_{- x} dx , x < 0
- du = ¹⁄_x dx , x < 0
For x > 0,
∫ ¹⁄_{x ln |x|} dx = ∫ ¹⁄_{ln |x|} ¹⁄_x dx
= ∫¹⁄_u du
= ln |u| + C
= ln |ln |x| | + C

For x < 0
∫ ¹⁄_{x ln |x|} dx = ∫ ¹⁄_{ln |x|} ¹⁄_x dx
= - ∫ ¹⁄_u du
= - ln |u| + C
= -ln |ln |x| | + C.

(d) sin⁵ x cos x; u = sin x
Solution
u = sin x
^du⁄_dx = cos x
du = cos x dx
∫ sin⁵ x cos x dx = ∫ u⁵ du
= ^u⁵⁺¹⁄₅₊₁ + C
= ^u⁶⁄₆ C
= ¹⁄₆ u⁶ + C
= ¹⁄₆ sin⁶ x + C.

(e) ^{ln x}⁄_x, x > 0; u = ln x
Solution
Let u = ln x
du = ¹⁄_x dx
∫ ^{ln x}⁄_x dx = ∫ ln x ¹⁄_x dx
= ∫ u du
= ^u¹⁺¹⁄₁₊₁ + C
= ^u²⁄₂ + C
= ¹⁄₂ u² + C.
= ¹⁄₂ (ln x)² + C.

(f) x³e^x⁴ ; u = x⁴
Solution
Let u = x⁴
du = 4x³ dx
¹⁄₄ du = x³ dx
∫ x³ e^x⁴ dx = ∫ e^x⁴ x³ dx
= ∫ e^u ¹⁄₄ du
= ¹⁄₄ ∫ e^u du
= ¹⁄₄ e^u + C
= ¹⁄₄ e^x⁴ + C .

2. Use the substitution method to evaluate the following integrals.
(a) ∫ x √ 1 - x dx
Solution
u = 1 - x ⇒ x = 1 - u
du = - dx
- du = dx

∫ x √ 1 - x dx = x (1 - x)^{^¹⁄₂} dx
= ∫ (1 - u) u^{^¹⁄₂} (- du)
= - ∫ (1 - u) u^{^¹⁄₂} du
= - ∫ [u^{^¹⁄₂} - u^{^³⁄₂}] du
= - [^{u^{^{¹⁄₂ + 1}}} ⁄ _{¹⁄₂ + 1} - ^{u^{^{³⁄₂ + 1}}} ⁄_{³⁄₂ + 1}] + C
= ^{u^{^⁵⁄₂}}⁄_⁵⁄₂ - ^{u^{^³⁄₂}}⁄_³⁄₂ + C
= ²⁄₅ (1 - x)^{^⁵⁄₂} - ²⁄₃ (1 - x)^{^³⁄₂} + C.

(b) ∫ (2x + 1)(x² + x)⁷ dx
Solution
Let u = x² + x
du = (2x + 1) dx
∫ (2x + 1)(x² + x)⁷ = ∫ (x² + x)⁷ (2x + 1) dx
= ∫ u⁷ du
= ^u⁷⁺¹⁄₇₊₁ + C
= ^u⁸⁄₈ + C
= ¹ ⁄₈ u⁸ + C
= ¹⁄₈ (x² + x)⁸ + C.

(c) ∫ sin³x dx
Solution
∫ sin³ x dx = sin² x sin x dx
= ∫ (1 - cos² x) sin x dx
Let u = cos x
du = - sin x dx
- du = sin x dx

∫ (1 - cos² x) sin x dx = ∫ (1 - u²(- du)
= ∫ (u² - 1) du
= [^u²⁺¹ ⁄_{2 + 1} - 1 ⋅ ^u⁰⁺¹ ⁄_{0 + 1}] + C
= ^u³⁄₃ - u + C
= ¹⁄₃ cos³ x - cos x + C.

(d) ∫ x² √ x³ - 2 dx
Solution
Let u = x³ - 2
du = 3x² dx
x² dx = ¹⁄₃ du
∫ x² √ x³ - 2 dx = ∫ √ x³ - 2 x² dx
= ∫ (x³ - 2)^{^¹⁄₂} x² dx
= ∫ u^{^¹⁄₂} ¹⁄₃ du
= ¹⁄₃ ∫ u^{^¹⁄₂} du
= ¹⁄₃ ⋅ ^{u^{^{¹⁄₂ + 1}}}⁄_{¹⁄₂ + 1} + C
= ¹⁄₃ ⋅ ^{u^{^³⁄₂}}⁄ _³⁄₂ + C
= ¹⁄₃ ⋅ ²⁄₃ u^{^³⁄₂} + C
= ²⁄₉ (x³ - 2) ^{^³⁄₂} + C

(e) ∫ ^{sec² x}⁄_{tan x} dx
Solution
Let u = tan x
^du⁄_dx = sec² x
du = sec² x dx

∫ ^{sec² x}⁄_{tan x} dx = ∫ ¹⁄_{tan x} sec² x dx
= ∫ ¹⁄_u du
= ln |u| + C
= ln |tan x| + C.

(f) ∫ ^x⁄_{√
x + 1} dx
Solution
Let u = x + 1 ⇒ x = u - 1
^du⁄_dx = 1
du = dx
∫ ^x⁄_{√
x + 1} dx = ∫ ^{u - 1}⁄_{√
u} du
= ∫ (^u⁄_{√
u} - ¹⁄_{√
u}) du
= ∫ (^{√
u ⋅
√
u}⁄_{√
u} - ¹⁄_{√
u}) du
= ∫ (u^{^¹⁄₂} - u^{^{- ¹⁄₂}}) du
= (^{u^{^{¹⁄₂ + 1}}}⁄_{¹⁄₂ + 1} - ^{u^{^{- ¹⁄₂ + 1}}}⁄_{- ¹⁄₂ + 1}) + C
= ^{u^{^³⁄₂}}⁄_³⁄₂ - ^{u^{^¹⁄₂}}⁄_¹⁄₂ + C
= ²⁄₃u^{^³⁄₂} - 2 u^{^¹⁄₂} + C.
= ²⁄₃(x + 1)^{^³⁄₂} - 2(x + 1)^{^¹⁄₂} + C.

3. Evaluate the integral ∫ ^x⁄_{(x² + 1) ln (x² + 1)} dx.
Solution
Let u = ln (x² + 1)
^du⁄_dx = ¹⁄_x² + 1 ⋅ ^d⁄_dx (x² + 1)
^du⁄_dx = ¹⁄_{x² + 1} ⋅ 2x
du = ¹⁄_{x² + 1} ⋅ 2x dx
¹⁄₂ du = ^x⁄_{x² + 1} ⋅ dx
∫ ^x⁄_{(x² + 1) ln (x² + 1)} dx = ∫ ¹⁄_{ln (x² + 1)} ⋅ ^x⁄_{x² + 1} dx
= ∫ ¹⁄_u ¹⁄₂ du
= ¹⁄₂ ∫ ¹⁄_u du
= ¹⁄₂ ln |u| + C
= ¹⁄₂ ln | ln (x² + 1)| + C.

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