Integrals · Lesson 2 of 9

Integration by Substitution

“Turn composite integrands into familiar forms by changing the variable carefully.”

Learning Objectives

• Recognise an inner function whose derivative appears in the integrand. • Transform the whole integrand and its differential into one new variable. • Use successive substitutions where needed and return to the original variable. • Derive the integrals of tangent, cotangent, secant, and cosecant. • Use algebra and trigonometric rewritings to reveal a substitution.

Reading the chain rule backwards

Inspection works when a derivative is easy to recognise. A composite expression can hide that derivative: in $2 x sin (x^{2} + 1)$ , the factor $2 x$ is the derivative of the inner expression $x^{2} + 1$ . Naming that inner expression changes the integral into a standard sine integral. The method is called substitution, or change of variable.

Suppose $x = g (t)$ . Then $d x = g' (t) d t$ , so every occurrence of $x$ , including its differential, must be transformed. Alternatively, when choosing $u = u (x)$ , write $d u = u' (x) d x$ . These are coordinated changes to the expression, not permission to discard unwanted factors.

Change of variableLaTeX

\int f (x) d x = \int f (g (t)) g' (t) d t

The right side is integrated with respect to

t

. For an indefinite integral, substitute back at the end so the result answers the original question.

This formula is supported by the chain rule. If $F' = f$ , then differentiating $F (g (t))$ gives $f (g (t)) g' (t)$ . Thus the transformed integrand has exactly the antiderivative expected from the original one. The derivative factor is what makes the method work.

Example — A constant chain factor

Problem
Find $\int sin (m x) d x$ for a nonzero constant $m$ .

1.Put $u = m x$ . Then $d u = m d x$ , so $d x = d u / m$ .
2.The integral becomes $(1/ m) \int sin u d u = - (1/ m) cos u + C$ .
3.Replace $u$ by $m x$ : the answer is $- cos (m x) / m + C$ . If $m = 0$ , the original integrand is zero and the answer is a constant instead.

Example — A visible inner derivative

Problem
Evaluate $\int 2 x sin (x^{2} + 1) d x$ .

1.Set $u = x^{2} + 1$ , so $d u = 2 x d x$ . The entire remaining expression becomes $sin u d u$ .
2.Integrate to obtain $- cos u + C$ .
3.Substitute back: $- cos (x^{2} + 1) + C$ . Its derivative is $2 x sin (x^{2} + 1)$ , including the required chain factor.

Example — Two substitutions working together

Problem
Evaluate $\int tan^{4} (x) sec^{2} (x) / x d x$ for $x > 0$ .

1.Put $t = x$ . Then $d x = 2 t d t$ , so the integral is $2 \int tan^{4} t sec^{2} t d t$ .
2.Now put $u = tan t$ , with $d u = sec^{2} t d t$ . The integral becomes $2 \int u^{4} d u = 2 u^{5} /5 + C$ .
3.Undo both changes: $\frac{2}{5} tan^{5} (x) + C$ . One can also choose $u = tan (x)$ directly; its derivative combines both changes.

Example — An inverse-trigonometric inner function

Problem
Evaluate $\int sin (tan^{- 1} x) / (1 + x^{2}) d x$ .

1.Choose $u = tan^{- 1} x$ , because $d u = d x / (1 + x^{2})$ .
2.The transformed integral is $\int sin u d u = - cos u + C$ .
3.The answer is $- cos (tan^{- 1} x) + C$ . An inverse trigonometric expression can be the inner function just as a polynomial can.

Four trigonometric integrals with logarithms

The remaining elementary trigonometric functions do not integrate by the power rule. Their formulas arise by rewriting each function as a quotient whose numerator is a derivative of its denominator. We use natural logarithms and absolute values because the functions can be positive or negative on different intervals.

Tangent and cotangentLaTeX

\int tan x d x = - ln ∣ cos x ∣ + C = ln ∣ sec x ∣ + C, \int cot x d x = ln ∣ sin x ∣ + C

These formulas hold on intervals avoiding the zeros of their denominators.

For tangent, write $tan x = sin x / cos x$ and set $u = cos x$ . Since $d u = - sin x d x$ , the integral is $- \int d u / u = - ln ∣ u ∣ + C$ . For cotangent, write $cos x / sin x$ and set $u = sin x$ ; now the differential has a positive sign, giving $ln ∣ u ∣ + C$ .

Secant and cosecantLaTeX

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

An equivalent cosecant answer is

- ln ∣ csc x + cot x ∣ + C

, because

(csc x - cot x) (csc x + cot x) = 1

For secant, multiply numerator and denominator by $sec x + tan x$ . The numerator becomes $sec x (sec x + tan x)$ , which is the derivative of the new denominator. Substituting that denominator reduces the integral to $\int d u / u$ . For cosecant, multiply by $(csc x + cot x) / (csc x + cot x)$ . The numerator is the negative of the denominator's derivative, so the result is $- ln ∣ csc x + cot x ∣ + C$ . The reciprocal identity explains the alternative formula above.

Rewriting before substituting

A useful substitution is sometimes concealed by an extra power, a shifted angle, or a numerator that is nearly a derivative. Before changing variables, look for a rewrite that creates a matching derivative. A few lines of algebra can make the actual integration short and transparent.

Example — Reserve one sine factor

Problem
Evaluate $\int sin^{3} x cos^{2} x d x$ .

1.Write $sin^{3} x cos^{2} x = (1 - cos^{2} x) cos^{2} x sin x$ . Reserving one sine factor prepares the derivative of cosine.
2.Put $u = cos x$ , so $d u = - sin x d x$ . The integral is $- \int (u^{2} - u^{4}) d u$ .
3.Integrate the polynomial and substitute back: $- cos^{3} x /3 + cos^{5} x /5 + C$ .

Example — A shifted trigonometric quotient

Problem
Evaluate $\int sin x / sin (x + a) d x$ , where $a$ is constant.

1.Let $u = x + a$ , so $x = u - a$ and $d x = d u$ . Expand $sin (u - a) = sin u cos a - cos u sin a$ .
2.Divide by $sin u$ : the new integrand is $cos a - sin a cot u$ .
3.Integration gives $u cos a - sin a ln ∣ sin u ∣ + C$ .
4.Replace $u$ by $x + a$ . The constant $a cos a$ is absorbed into $C$ , leaving $x cos a - sin a ln ∣ sin (x + a) ∣ + C$ on an interval where the original quotient exists.

Example — Create a denominator derivative

Problem
Evaluate $\int d x / (1 + tan x)$ .

1.Where the original expression is defined, rewrite it as $cos x / (cos x + sin x)$ .
2.Split the numerator: $cos x = \frac{1}{2} (cos x + sin x) + \frac{1}{2} (cos x - sin x)$ .
3.The first resulting integral is $x /2$ . In the second, choose $u = cos x + sin x$ , whose derivative is $cos x - sin x$ .
4.The answer is $x /2 + \frac{1}{2} ln ∣ cos x + sin x ∣ + C$ . The rewrite does not extend the domain of the original expression through a point where tangent is undefined.

Example — A logarithmic inner function

Problem
Evaluate $\int (ln x)^{2} / x d x$ for $x > 0$ .

1.Choose $u = ln x$ because $d u = d x / x$ .
2.Integrate $\int u^{2} d u = u^{3} /3 + C$ .
3.Therefore the answer is $(ln x)^{3} /3 + C$ . The factor $1/ x$ is essential; without it this substitution would not give the same simple power integral.

Common mistake

After substitution, keep only one integration variable in the transformed expression. For example, $u = x^{2} + 1$ changes $2 x d x$ to $d u$ , not $d x$ alone. Also return to $x$ after an indefinite integral; a result in an undefined temporary variable is incomplete.

Quick check

For $\int 2 x e^{x^{2}} d x$ , which substitution immediately removes the derivative factor?

Quick check

What is $\int 3 cos (3 x) d x$ ?

Quick check

If $u = ln x$ , which differential is correct?

Quick check

Which is $\int tan x d x$ ?

Quick check

Why does $u = cos x$ help with $\int sin^{3} x cos^{2} x d x$ ?

Practice Problems

Find $\int 2 x / (1 + x^{2}) d x$ .
Find $\int d x / [x (1 + ln x)]$ on an appropriate interval.
Find $\int (4 x + 2) x^{2} + x + 1 d x$ .
Find $\int e^{2 x} / (e^{2 x} + 1) d x$ .
Find $\int sin^{- 1} x / 1 - x^{2} d x$ .
Find $\int cot x ln ∣ sin x ∣ d x$ .
Find $\int d x / (1 + cot x)$ and explain how it relates to the worked tangent-quotient example.
Find $\int (x + 1) (x + ln x)^{2} / x d x$ for $x > 0$ .

Key Takeaways

Substitution reverses the chain rule by changing both the expression and its differential. Choose an inner function whose derivative is present or can be created by a rewrite. All terms in the transformed integral must use the same variable. Tangent, cotangent, secant, and cosecant have logarithmic antiderivatives. Substitute back and differentiate the final answer to check it.

Previous · Lesson 1

Antiderivatives and Indefinite Integrals

Next · Lesson 3

Integrating with Trigonometric Identities

Reading the chain rule backwards

Four trigonometric integrals with logarithms

Rewriting before substituting

Quiz

Practice Problems

Key Takeaways