Reverse Chain Rule

Integrating composite functions using the "guess and adjust" method

SPEC 8.5 • INTEGRATION

The Reverse Chain Rule

When we differentiate a composite function using the chain rule, we get:

Chain Rule

\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)

The reverse chain rule allows us to integrate expressions that are the result of chain rule differentiation.

Key Recognition

Look for integrands of the form: f'(g(x)) × g'(x)

The derivative of the inner function must appear as a factor (possibly with a constant multiple).

Strategy

We use a "guess and adjust" method: guess what the answer might be, differentiate to check, then adjust by a constant factor if needed.

The "Guess and Adjust" Method

4-Step Method

GUESS: Let y = [likely answer] — raise the power, or use appropriate form

DIFFERENTIATE: Find dy/dx using the chain rule

COMPARE: Check what coefficient you get versus what you need

ADJUST: Multiply or divide by the appropriate constant

Why This Works

Integration is the reverse of differentiation. If we know what differentiates to give our integrand (up to a constant), we can find the integral by adjusting for that constant.

Important

This method only works when the derivative of the inner function appears as a constant multiple in the integrand. If there's a variable mismatch, we need other techniques.

Linear Functions - Quick Formula

For linear inner functions (ax + b), there's a direct formula:

Linear Case Formula

\int (a x + b)^{n} d x = \frac{( a x + b ) ^{n + 1}}{a ( n + 1 )} + C

For polynomial inner functions like (x² + k) or (x³ + k), we must check that the derivative of the inner function appears:

Polynomial Inner Functions

For

\int x \cdot f (x^{2}) d x

: the x is "part of" the derivative of x² (which is 2x)

For

\int x^{2} \cdot f (x^{3}) d x

: the x² is "part of" the derivative of x³ (which is 3x²)

Pattern

The power of x outside must be exactly one less than the power inside the composite function.

Negative and Fractional Powers

The same method works for negative powers (fractions) and fractional powers (roots):

Negative Powers

\int \frac{x}{( x ^{2} + 1 ) ^{2}} d x = \int x (x^{2} + 1)^{- 2} d x

Guess: y = (x²+1)⁻¹ = 1/(x²+1)

Fractional Powers (Roots)

\int \frac{x}{x ^{2} + 3} d x = \int x (x^{2} + 3)^{- 1/2} d x

Guess: y = (x²+3)^(1/2) = √(x²+3)

Key Insight

For negative powers: raise the power by 1 (e.g., −2 → −1 or −3 → −2)

For roots in denominator: the integral will have a positive power (e.g., −1/2 → +1/2)

The Logarithmic Form

A special case occurs when the numerator is the derivative of the denominator:

Key Result

\int \frac{f ^{'} ( x )}{f ( x )} d x = ln ∣ f (x) ∣ + C

This works because differentiating ln|f(x)| gives f'(x)/f(x) by the chain rule.

How to Recognise

Look for a fraction where:

The denominator is a function f(x)

The numerator is (a constant multiple of) the derivative f'(x)

Examples

\int \frac{2 x}{x ^{2} + 1} d x = ln ∣ x^{2} + 1∣ + C

[numerator 2x is derivative of x² + 1]

\int \frac{c o s x}{s i n x} d x = ln ∣ sin x ∣ + C

[cos x is derivative of sin x]

Remember

Always include the modulus signs: ln|f(x)| because logarithms are only defined for positive values.

Special Trigonometric Integrals

Using the

\frac{f ^{'} ( x )}{f ( x )}

pattern, we get these important results:

Key Results

\int tan x d x = - ln ∣ cos x ∣ + C

\int cot x d x = ln ∣ sin x ∣ + C

\int sec x d x = ln ∣ sec x + tan x ∣ + C

\int csc x d x = - ln ∣ csc x + cot x ∣ + C

tan x

Let y = ln|cos x|

dy/dx = (−sin x)/cos x = −tan x

∴ ∫tan x dx = −ln|cos x| + C

cot x

Let y = ln|sin x|

dy/dx = cos x/sin x = cot x

∴ ∫cot x dx = ln|sin x| + C

sec x

Let y = ln|sec x + tan x|

dy/dx = (sec x tan x + sec²x)/(sec x + tan x) = sec x

∴ ∫sec x dx = ln|sec x + tan x| + C

csc x

Let y = ln|csc x + cot x|

dy/dx = (−csc x cot x − csc²x)/(csc x + cot x) = −csc x

∴ ∫csc x dx = −ln|csc x + cot x| + C

Worked Example — Linear Power01

Find $\int (3 x + 2)^{4} d x$

1

GUESS: Let y = (3x + 2)⁵

2

DIFFERENTIATE: dy/dx = 5(3x + 2)⁴ × 3 = 15(3x + 2)⁴

3

COMPARE: We get 15(3x + 2)⁴ but we want 1(3x + 2)⁴

4

ADJUST: Divide by 15

Answer

\frac{( 3 x + 2 ) ^{5}}{15} + C

Worked Example — Linear (Negative Coefficient)02

Find $\int (5 - 2 x)^{6} d x$

1

GUESS: Let y = (5 − 2x)⁷

2

DIFFERENTIATE: dy/dx = 7(5 − 2x)⁶ × (−2) = −14(5 − 2x)⁶

3

COMPARE: We get −14(5 − 2x)⁶ but we want 1(5 − 2x)⁶

4

ADJUST: Divide by −14

Answer

- \frac{( 5 - 2 x ) ^{7}}{14} + C

Worked Example — Polynomial Inner03

Find $\int 6 x (x^{2} + 5)^{3} d x$

1

GUESS: Let y = (x² + 5)⁴

2

DIFFERENTIATE: dy/dx = 4(x² + 5)³ × 2x = 8x(x² + 5)³

3

COMPARE: We get 8x(x² + 5)³ but we want 6x(x² + 5)³

4

ADJUST: Multiply by 6/8 = 3/4

Answer

\frac{3 ( x ^{2} + 5 ) ^{4}}{4} + C

Worked Example — Polynomial Inner (Cubic)04

Find $\int x^{2} (x^{3} - 1)^{5} d x$

1

GUESS: Let y = (x³ − 1)⁶

2

DIFFERENTIATE: dy/dx = 6(x³ − 1)⁵ × 3x² = 18x²(x³ − 1)⁵

3

COMPARE: We get 18x²(x³ − 1)⁵ but we want 1x²(x³ − 1)⁵

4

ADJUST: Divide by 18

Answer

\frac{( x ^{3} - 1 ) ^{6}}{18} + C

Worked Example — Trigonometric (Linear)05

Find $\int sin (5 x) d x$

1

GUESS: Let y = −cos(5x) [since d/dx(−cos x) = sin x]

2

DIFFERENTIATE: dy/dx = −(−sin(5x)) × 5 = 5sin(5x)

3

COMPARE: We get 5sin(5x) but we want 1sin(5x)

4

ADJUST: Divide by 5

Answer

- \frac{c o s ( 5 x )}{5} + C

Worked Example — Trig with Polynomial06

Find $\int x^{2} cos (x^{3}) d x$

1

GUESS: Let y = sin(x³) [since d/dx(sin u) = cos u]

2

DIFFERENTIATE: dy/dx = cos(x³) × 3x² = 3x²cos(x³)

3

COMPARE: We get 3x²cos(x³) but we want 1x²cos(x³)

4

ADJUST: Divide by 3

Answer

\frac{s i n ( x ^{3} )}{3} + C

Worked Example — sec² (Linear)07

Find $\int sec^{2} (4 x) d x$

1

GUESS: Let y = tan(4x) [since d/dx(tan x) = sec²x]

2

DIFFERENTIATE: dy/dx = sec²(4x) × 4 = 4sec²(4x)

3

COMPARE: We get 4sec²(4x) but we want 1sec²(4x)

4

ADJUST: Divide by 4

Answer

\frac{t a n ( 4 x )}{4} + C

Worked Example — Trig Powers (sin³x)08

Find $\int sin^{3} (x) cos (x) d x$

1

GUESS: Let y = sin⁴(x) [raise power of sin by 1]

2

DIFFERENTIATE: dy/dx = 4sin³(x) × cos(x) = 4sin³(x)cos(x)

3

COMPARE: We get 4sin³(x)cos(x) but we want 1sin³(x)cos(x)

4

ADJUST: Divide by 4

Answer

\frac{s i n ^{4} ( x )}{4} + C

Worked Example — Trig Powers (cos⁴x)09

Find $\int cos^{4} (x) sin (x) d x$

1

GUESS: Let y = cos⁵(x) [raise power of cos by 1]

2

DIFFERENTIATE: dy/dx = 5cos⁴(x) × (−sin(x)) = −5cos⁴(x)sin(x)

3

COMPARE: We get −5cos⁴(x)sin(x) but we want +1cos⁴(x)sin(x)

4

ADJUST: Divide by −5

Answer

- \frac{c o s ^{5} ( x )}{5} + C

10

Worked Example — Trig Powers (tan × sec²)

Find $\int tan (x) sec^{2} (x) d x$

1

GUESS: Let y = tan²(x) [since d/dx(tan x) = sec²x]

2

DIFFERENTIATE: dy/dx = 2tan(x) × sec²(x) = 2tan(x)sec²(x)

3

COMPARE: We get 2tan(x)sec²(x) but we want 1tan(x)sec²(x)

4

ADJUST: Divide by 2

Answer

\frac{t a n ^{2} ( x )}{2} + C

11

Worked Example — Exponential

Find $\int 4 x \cdot e^{x^{2}} d x$

1

GUESS: Let y = e^(x²)

2

DIFFERENTIATE: dy/dx = e^(x²) × 2x = 2x·e^(x²)

3

COMPARE: We get 2x·e^(x²) but we want 4x·e^(x²)

4

ADJUST: Multiply by 2

Answer

2 e^{x^{2}} + C

12

Worked Example — Logarithmic

Find $\int \frac{6 x ^{2}}{x ^{3} + 4} d x$

1

GUESS: Let y = ln|x³ + 4| [since d/dx(ln u) = u'/u]

2

DIFFERENTIATE: dy/dx = 3x²/(x³ + 4)

3

COMPARE: We get 3x²/(x³+4) but we want 6x²/(x³+4)

4

ADJUST: Multiply by 2

Answer

2 ln ∣ x^{3} + 4∣ + C

13

Worked Example — Logarithmic (Trig)

Find $\int \frac{c o s x}{s i n x} d x$

1

RECOGNISE: Numerator (cos x) is the derivative of denominator (sin x)

2

GUESS: Let y = ln|sin x|

3

DIFFERENTIATE: dy/dx = cos x / sin x — perfect match!

4

ADJUST: No adjustment needed

Note: This can also be written as −ln|csc x| + C

Answer

ln ∣ sin x ∣ + C

14

Worked Example — Logarithmic (tan x)

Find $\int tan (x) d x$

1

REWRITE:

\int tan (x) d x = \int \frac{s i n x}{c o s x} d x

2

RECOGNISE: Numerator (sin x) is almost the derivative of denominator (cos x) — but d/dx(cos x) = −sin x

3

GUESS: Let y = ln|cos x|

4

DIFFERENTIATE: dy/dx = (−sin x)/cos x = −tan x

5

ADJUST: Multiply by −1

Note: This can also be written as ln|sec x| + C

Answer

- ln ∣ cos x ∣ + C

15

Worked Example — Negative Power (Fraction)

Find $\int \frac{x}{( x ^{2} + 1 ) ^{2}} d x$

1

REWRITE:

\int x (x^{2} + 1)^{- 2} d x

[power is −2]

2

GUESS: Let y = (x² + 1)⁻¹ = 1/(x² + 1) [raise power: −2 → −1]

3

DIFFERENTIATE: dy/dx = −1(x² + 1)⁻² × 2x = −2x/(x² + 1)²

4

COMPARE: We get −2x/(x²+1)² but we want +1x/(x²+1)²

5

ADJUST: Divide by −2

Answer

- \frac{1}{2 ( x ^{2} + 1 )} + C

16

Worked Example — Negative Power (Cubic)

Find $\int \frac{x ^{2}}{( x ^{3} - 2 ) ^{3}} d x$

1

REWRITE:

\int x^{2} (x^{3} - 2)^{- 3} d x

[power is −3]

2

GUESS: Let y = (x³ − 2)⁻² [raise power: −3 → −2]

3

DIFFERENTIATE: dy/dx = −2(x³ − 2)⁻³ × 3x² = −6x²/(x³ − 2)³

4

COMPARE: We get −6x²/(x³−2)³ but we want +1x²/(x³−2)³

5

ADJUST: Divide by −6

Answer

- \frac{1}{6 ( x ^{3} - 2 ) ^{2}} + C

17

Worked Example — Square Root Denominator

Find $\int \frac{x}{x ^{2} + 3} d x$

1

REWRITE:

\int x (x^{2} + 3)^{- 1/2} d x

[power is −1/2]

2

GUESS: Let y = (x² + 3)^(1/2) = √(x² + 3) [raise: −1/2 → +1/2]

3

DIFFERENTIATE: dy/dx = (1/2)(x² + 3)^(−1/2) × 2x = x/√(x² + 3)

4

COMPARE: We get x/√(x²+3) and we want x/√(x²+3) — perfect match!

5

ADJUST: No adjustment needed (multiply by 1)

Answer

x^{2} + 3 + C

18

Worked Example — Trig Square Root

Find $\int \frac{c o s x}{s i n x} d x$

1

REWRITE:

\int cos (x) \cdot [sin (x)]^{- 1/2} d x

[power is −1/2]

2

GUESS: Let y = [sin(x)]^(1/2) = √(sin x) [raise: −1/2 → +1/2]

3

DIFFERENTIATE: dy/dx = (1/2)[sin(x)]^(−1/2) × cos(x) = cos(x)/(2√(sin x))

4

COMPARE: We get (1/2)cos(x)/√(sin x) but we want 1·cos(x)/√(sin x)

5

ADJUST: Multiply by 2

Answer

2 sin x + C

Practice

Question 1: Find

\int (2 x - 7)^{5} d x

GUESS: Let y = (2x − 7)⁶

DIFFERENTIATE: dy/dx = 6(2x − 7)⁵ × 2 = 12(2x − 7)⁵

ADJUST: Divide by 12

Answer

\frac{( 2 x - 7 ) ^{6}}{12} + C

Practice

Question 2: Find

\int sin^{5} (x) cos (x) d x

GUESS: Let y = sin⁶(x)

DIFFERENTIATE: dy/dx = 6sin⁵(x) × cos(x) = 6sin⁵(x)cos(x)

ADJUST: Divide by 6

Answer

\frac{s i n ^{6} ( x )}{6} + C

Practice

Question 3: Find

\int \frac{x ^{3}}{( x ^{4} + 2 ) ^{2}} d x

REWRITE:

\int x^{3} (x^{4} + 2)^{- 2} d x

GUESS: Let y = (x⁴ + 2)⁻¹ = 1/(x⁴ + 2)

DIFFERENTIATE: dy/dx = −1(x⁴ + 2)⁻² × 4x³ = −4x³/(x⁴ + 2)²

ADJUST: Divide by −4

Answer

- \frac{1}{4 ( x ^{4} + 2 )} + C

Summary

The 4-Step Method

1

GUESS — Let y = [likely answer]

2

DIFFERENTIATE — Find dy/dx using chain rule

3

COMPARE — Check coefficient

4

ADJUST — Multiply/divide as needed

Common Guesses

Integrand TypeGuess y = ...

[f (x)]^{n}

[f (x)]^{n + 1}

sin [f (x)]

- cos [f (x)]

cos [f (x)]

sin [f (x)]

sin^{n} (x) cos (x)

sin^{n + 1} (x)

cos^{n} (x) sin (x)

cos^{n + 1} (x)

sec^{2} [f (x)]

tan [f (x)]

e^{f (x)}

e^{f (x)}

\frac{f ^{'} ( x )}{f ( x )}

ln ∣ f (x) ∣

\frac{1}{[ f ( x ) ] ^{n}}

[f (x)]^{- (n - 1)}

\frac{1}{f ( x )}

f (x)