MathsCBSE Class 12 · Integrals

Integration by Substitution: Choosing t and Changing Limits (Class 12)

Updated 15 October 2026 · Fact-checked

Integration by substitution reverses the chain rule. If the integrand looks like f(g(x)) g′(x), put t = g(x), so dt = g′(x) dx, and the integral becomes ∫f(t) dt. Integrate in t and substitute back. In a definite integral, change the limits to t-values instead of substituting back.

Pick t = g(x), replace g′(x)dx by dt, integrate in t, then go back to x; for definite integrals change the limits instead.

Illustration drawn for QuestionBuddy.

Understand Integration by Substitution

The chain rule says $\frac{d}{d x} F (g (x)) = F^{'} (g (x)) g^{'} (x)$ . Reading it backwards gives the substitution rule:
$\int f (g (x)) g^{'} (x) d x = \int f (t) d t$ , where $t = g (x)$ and $d t = g^{'} (x) d x$ .

**How to choose $t$ .** Look for a function whose derivative is also present (up to a constant factor):

$\frac{( l o g x ) ^{2}}{x}$ : put $t = lo g x$ , since $\frac{1}{x} d x$ is there.
$e^{t a n x} sec^{2} x$ : put $t = tan x$ .
$\frac{c o s x}{x}$ : put $t = x$ , so $d t = \frac{d x}{2 x}$ .
$\frac{f ^{'} ( x )}{f ( x )}$ : put $t = f (x)$ , giving $lo g ∣ f (x) ∣ + C$ .
an inner expression in a bracket or root, like $(1 + x^{3})$ in $\frac{x ^{5}}{1 + x ^{3}}$ , often works even if extra factors remain; write the leftover $x$ 's in terms of $t$ .

If the derivative of your $t$ appears with a different constant, adjust: $\int x e^{x^{2}} d x$ with $t = x^{2}$ , $d t = 2 x d x$ , gives $\frac{1}{2} \int e^{t} d t$ .

Trigonometric substitutions and identities. Sometimes a clever division makes $t$ appear. For $\frac{s i n 2 x}{s i n ^{4} x + c o s ^{4} x}$ , divide numerator and denominator by $cos^{4} x$ to get $\frac{2 t a n x s e c ^{2} x}{1 + t a n ^{4} x}$ ; then $t = tan^{2} x$ turns it into $\int \frac{d t}{1 + t ^{2}}$ . For expressions like $\frac{x ^{2} - 1}{x ^{4} + x ^{2} + 1}$ , divide by $x^{2}$ and put $t = x + \frac{1}{x}$ .

Standard results often reached after substitution (from the syllabus list): $\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a}$ , $\int \frac{d x}{x ^{2} - a ^{2}} = \frac{1}{2 a} lo g \frac{x - a}{x + a}$ , $\int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} \frac{x}{a}$ , $\int \frac{d x}{x ^{2} \pm a ^{2}} = lo g x + x^{2} \pm a^{2}$ , each $+ C$ .

Definite integrals by substitution. When $x$ runs from $a$ to $b$ , $t = g (x)$ runs from $g (a)$ to $g (b)$ . Change the limits along with the variable and never return to $x$ :
$\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (t) d t$ .
For example, $\int_{0}^{π /2} \frac{s i n x}{1 + c o s ^{2} x} d x$ with $t = cos x$ becomes $\int_{1}^{0} \frac{- d t}{1 + t ^{2}} = \int_{0}^{1} \frac{d t}{1 + t ^{2}} = \frac{π}{4}$ .

Always check an indefinite answer by differentiating it. Substitution is NCERT Section 7.3 (with definite integrals by substitution in 7.9) and appears in every board paper.

Integration by Substitution: formulas to remember

Substitution rule: $\int f (g (x)) g^{'} (x) d x = \int f (t) d t, t = g (x)$; Reverse of the chain rule.
Log rule: $\int \frac{f ^{'} ( x )}{f ( x )} d x = lo g ∣ f (x) ∣ + C$; Numerator is the derivative of the denominator.
Power of a function: $\int [f (x)]^{n} f^{'} (x) d x = \frac{[ f ( x ) ] ^{n + 1}}{n + 1} + C, n \neq = - 1$; For example $\int \frac{( l o g x ) ^{2}}{x} d x = \frac{( l o g x ) ^{3}}{3} + C$ .
Standard forms: $\int \frac{d x}{x ^{2} + a ^{2}} = \frac{1}{a} tan^{- 1} \frac{x}{a} + C, \int \frac{d x}{a ^{2} - x ^{2}} = sin^{- 1} \frac{x}{a} + C$; Often the target after substitution.
Definite integral with substitution: $\int_{a}^{b} f (g (x)) g^{'} (x) d x = \int_{g (a)}^{g (b)} f (t) d t$; Change the limits; do not substitute back.
Exponential and trig: $\int e^{f (x)} f^{'} (x) d x = e^{f (x)} + C, \int sin (f (x)) f^{'} (x) d x = - cos (f (x)) + C$; Same pattern for any standard function.

How to solve Integration by Substitution questions

Use this routine for any integral that is not directly standard.

1Look for an inner function whose derivative (up to a constant) multiplies the rest; call it $t$ .
2Write $d t = g^{'} (x) d x$ and solve for the piece that appears in the integral (for example $x d x = \frac{d t}{2}$ ).
3Replace every $x$ in the integrand by an expression in $t$ ; no $x$ may remain.
4Integrate in $t$ using standard results.
5Indefinite integral: substitute $t = g (x)$ back and add $+ C$ . Definite integral: change the limits to $t$ -values and evaluate directly.
6Check an indefinite answer by differentiating it.

Integration by Substitution in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Distance from a speed formula

Suppose a cyclist's speed, $t$ hours after starting, is $v (t) = \frac{8 t}{1 + t ^{2}}$ km/h.

How far does the cyclist travel in the first $3$ hours?

Try it first, then show the solutionSolution

Distance $= \int_{0}^{3} \frac{8 t}{1 + t ^{2}} d t$ .
Put $u = 1 + t^{2}$ , $d u = 2 t d t$ ; limits: $t = 0 \to u = 1$ , $t = 3 \to u = 4$ .
$= \int_{1}^{4} \frac{4}{u} d u = [8 u]_{1}^{4} = 8 (2 - 1) = 8$ km.

Answer: 8 km

Memory hook: Distance is the integral of speed; substitution handles the root.

Total from a rate with a trigonometric substitution

Imagine a hostel's water use during the 12 daytime hours follows $w (t) = 50 sin (\frac{π t}{12})$ litres per hour, $0 \leq t \leq 12$ .

Find the total water used in these 12 hours.

Try it first, then show the solutionSolution

Total $= \int_{0}^{12} 50 sin \frac{π t}{12} d t$ .
Put $u = \frac{π t}{12}$ , $d t = \frac{12}{π} d u$ ; limits $0 \to 0$ and $12 \to π$ .
$= \frac{600}{π} \int_{0}^{π} sin u d u = \frac{600}{π} [- cos u]_{0}^{π} = \frac{600}{π} \times 2 = \frac{1200}{π}$ .
$\frac{1200}{π} \approx 382$ litres.

Answer: $\frac{1200}{π} \approx 382$ litres

Memory hook: Rate in, total out: integrate, and let a substitution tidy the angle.

Tricks to remember Integration by Substitution

Short memory aids. Say each one aloud once, then use it on a question the same day.

Find the function and its derivative
Scan for a pair: a function inside and its derivative outside. That function is your $t$ .
Use it when: Choosing the substitution.
No x left behind
After substituting, the integral must contain only $t$ and $d t$ . Leftover $x$ 's must be rewritten in $t$ .
Use it when: Integrals like $\frac{x ^{5}}{1 + x ^{3}}$ .
Change the limits, change the passport
When $x$ becomes $t$ , the limits must become $t$ -values too.
Use it when: Definite integrals.
Differentiate to check
Your answer is right if its derivative gives back the integrand.
Use it when: A 30-second final check.

Integration by Substitution: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: A power of log x · 2 marks · Very short answerBoard level

Evaluate $\int \frac{( l o g x ) ^{2}}{x} d x$ .

Try it first, then show the solutionSolution

Put $t = lo g x$ , so $d t = \frac{1}{x} d x$ . (1 mark)
$\int t^{2} d t = \frac{t ^{3}}{3} + C = \frac{( l o g x ) ^{3}}{3} + C$ . (1 mark)

Answer: $\frac{( l o g x ) ^{3}}{3} + C$

Example 2: A root inside a cosine · 2 marks · Very short answerBoard level

Evaluate $\int \frac{c o s x}{x} d x$ .

Try it first, then show the solutionSolution

Put $t = x$ , $d t = \frac{d x}{2 x}$ , so $\frac{d x}{x} = 2 d t$ . (1 mark)
$\int 2 cos t d t = 2 sin t + C = 2 sin x + C$ . (1 mark)

Answer: $2 sin x + C$

Example 3: Divide first, then substitute · 3 marks · Short answerHarder board level

Evaluate $\int \frac{s i n 2 x}{s i n ^{4} x + c o s ^{4} x} d x$ .

Try it first, then show the solutionSolution

Write $sin 2 x = 2 sin x cos x$ and divide numerator and denominator by $cos^{4} x$ : integrand $= \frac{2 t a n x s e c ^{2} x}{t a n ^{4} x + 1}$ . (1 mark)
Put $t = tan^{2} x$ , $d t = 2 tan x sec^{2} x d x$ . (1 mark)
$\int \frac{d t}{1 + t ^{2}} = tan^{- 1} t + C = tan^{- 1} (tan^{2} x) + C$ . (1 mark)

Answer: $tan^{- 1} (tan^{2} x) + C$

Example 4: A definite integral with changed limits · 3 marks · Short answerHarder board level

Evaluate $\int_{0}^{π /2} \frac{s i n x}{1 + c o s ^{2} x} d x$ .

Try it first, then show the solutionSolution

Put $t = cos x$ , $d t = - sin x d x$ . Limits: $x = 0 \to t = 1$ ; $x = \frac{π}{2} \to t = 0$ . (1 mark)
$\int_{1}^{0} \frac{- d t}{1 + t ^{2}} = \int_{0}^{1} \frac{d t}{1 + t ^{2}}$ . (1 mark)
$= [tan^{- 1} t]_{0}^{1} = \frac{π}{4}$ . (1 mark)

Answer: $\frac{π}{4}$

Example 5: Leftover x written in terms of t · 3 marks · Short answerHOTS / competency

Evaluate $\int \frac{x ^{5}}{1 + x ^{3}} d x$ .

Try it first, then show the solutionSolution

Put $t = 1 + x^{3}$ , $d t = 3 x^{2} d x$ , and $x^{3} = t - 1$ . Then $x^{5} d x = x^{3} \cdot x^{2} d x = (t - 1) \frac{d t}{3}$ . (1 mark)
$\frac{1}{3} \int \frac{t - 1}{t} d t = \frac{1}{3} \int (t^{1/2} - t^{- 1/2}) d t = \frac{1}{3} (\frac{2}{3} t^{3/2} - 2 t^{1/2}) + C$ . (1 mark)
$= \frac{2}{9} (1 + x^{3})^{3/2} - \frac{2}{3} (1 + x^{3})^{1/2} + C$ . (1 mark)

Answer: $\frac{2}{9} (1 + x^{3})^{3/2} - \frac{2}{3} 1 + x^{3} + C$

Example 6: Case study: absorption of a medicine · 4 marks · Case-basedHOTS / competency

Suppose a medicine enters the bloodstream at the rate $r (t) = 6 t e^{- t^{2}}$ mg per hour, $t$ hours after it is taken.
(i) Find $\int 6 t e^{- t^{2}} d t$ . (1 mark)
(ii) How much is absorbed in the first hour? (1 mark)
(iii) How much is absorbed between $t = 1$ and $t = 2$ ? (1 mark)
(iv) What total does the absorption approach after a long time? (1 mark)

Try it first, then show the solutionSolution

(i) Put $u = - t^{2}$ , $d u = - 2 t d t$ : $\int 6 t e^{- t^{2}} d t = - 3 e^{- t^{2}} + C$ . (1 mark)
(ii) $[- 3 e^{- t^{2}}]_{0}^{1} = 3 (1 - e^{- 1}) \approx 1.90$ mg. (1 mark)
(iii) $[- 3 e^{- t^{2}}]_{1}^{2} = 3 (e^{- 1} - e^{- 4}) \approx 1.05$ mg. (1 mark)
(iv) Over $[0, T]$ the total is $3 (1 - e^{- T^{2}})$ , which approaches 3 mg as $T$ grows. (1 mark)

Answer: (i) $- 3 e^{- t^{2}} + C$ (ii) about 1.90 mg (iii) about 1.05 mg (iv) 3 mg.

Where students lose marks in Integration by Substitution

The slips examiners see most often, and what to write instead.

Mistake: Leaving x in the integral after substituting
An integral in mixed variables cannot be evaluated.
Wrong working
$t = 1 + x^{3}$ : $\int \frac{x ^{5}}{t} \cdot \frac{d t}{3 x ^{2}}$ , then integrate treating $x$ as constant.
Right working
Write $x^{3} = t - 1$ : $\frac{1}{3} \int \frac{t - 1}{t} d t$ .
Do this: Express every remaining $x$ in terms of $t$ .
Usually costs: Most of the marks.
Mistake: Not changing the limits
The old limits belong to $x$ , not $t$ .
Wrong working
$\int_{0}^{π /2} \frac{s i n x}{1 + c o s ^{2} x} d x = \int_{0}^{π /2} \frac{- d t}{1 + t ^{2}}$ .
Right working
$= \int_{1}^{0} \frac{- d t}{1 + t ^{2}} = \frac{π}{4}$ .
Do this: Write the new limits next to the substitution line.
Usually costs: The value.
Mistake: Forgetting the constant factor in dt
$d t = 2 x d x$ means $x d x = \frac{d t}{2}$ .
Wrong working
$\int x e^{x^{2}} d x = e^{x^{2}} + C$ .
Right working
$\int x e^{x^{2}} d x = \frac{1}{2} e^{x^{2}} + C$ .
Do this: Solve $d t = \dots$ for exactly the piece in the integrand.
Usually costs: Half to one mark.
Mistake: Omitting + C
An indefinite integral is a family of functions.
Wrong working
$\int \frac{( l o g x ) ^{2}}{x} d x = \frac{( l o g x ) ^{3}}{3}$ .
Right working
$= \frac{( l o g x ) ^{3}}{3} + C$ .
Do this: Write $+ C$ in every indefinite answer.
Usually costs: Half a mark.

Exam tips

Substitution questions appeared in every board paper from 2023 to 2026, as MCQs and 2- or 3-mark questions; show the substitution line ' $t = \dots$ , $d t = \dots$ '.
In definite integrals, change the limits and say so; returning to $x$ is allowed but takes longer.
If the first substitution fails, try dividing by a power of $cos x$ or $x$ to create the derivative you need.
Keep standard forms (like $\int \frac{d x}{x ^{2} + a ^{2}}$ ) on your fingertips; substitution usually leads to one of them.
Check indefinite answers by differentiating; it catches missing constants.

Practice questions on Integration by Substitution

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
$\int \frac{e ^{x}}{1 + e ^{x}} d x$ equals:
(a) $e^{x} + C$
(b) $lo g (1 + e^{x}) + C$
(c) $tan^{- 1} e^{x} + C$
(d) $\frac{1}{1 + e ^{x}} + C$
Hint
The numerator is the derivative of the denominator.
Answer
(b) $lo g (1 + e^{x}) + C$ .
Q2 · 2 marks · Very short answerBoard level
Evaluate $\int \frac{s e c ^{2} ( l o g x )}{x} d x$ .
Hint
Put $t = lo g x$ .
Answer
$\int sec^{2} t d t = tan t + C = tan (lo g x) + C$ .
Q3 · 2 marks · Very short answerBoard level
Evaluate $\int \frac{x}{x + 4} d x$ .
Hint
Put $t = x + 4$ , so $x = t - 4$ .
Answer
$\int \frac{t - 4}{t} d t = \frac{2}{3} t^{3/2} - 8 t^{1/2} + C = \frac{2}{3} (x + 4)^{3/2} - 8 x + 4 + C$ .
Q4 · 3 marks · Short answerHarder board level
Evaluate $\int \frac{d x}{x ( 1 + x )}$ .
Hint
Put $t = x$ , so $x = t^{2}$ .
Answer
$t = x$ , $d t = \frac{d x}{2 x}$ : $\int \frac{2 d t}{1 + t ^{2}} = 2 tan^{- 1} t + C = 2 tan^{- 1} x + C$ .
Q5 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $\int 2 x cos (x^{2}) d x = sin (x^{2}) + C$ .
Reason (R): $\frac{d}{d x} sin (x^{2}) = 2 x cos (x^{2})$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Integration reverses differentiation.
Answer
(a). The derivative in R is exactly the integrand in A, which is why A holds.
Q6 · 3 marks · Short answerHarder board level
Evaluate $\int_{0}^{1} \frac{x}{1 + x ^{4}} d x$ .
Hint
Put $t = x^{2}$ and change the limits.
Answer
$t = x^{2}$ , $d t = 2 x d x$ , limits 0 to 1: $\frac{1}{2} \int_{0}^{1} \frac{d t}{1 + t ^{2}} = \frac{1}{2} \cdot \frac{π}{4} = \frac{π}{8}$ .
Q7 · 3 marks · Short answerHOTS / competency
Evaluate $\int \frac{x ^{2} - 1}{x ^{4} + x ^{2} + 1} d x$ .
Hint
Divide numerator and denominator by $x^{2}$ and put $t = x + \frac{1}{x}$ .
Answer
Integrand $= \frac{1 - \frac{1}{x ^{2}}}{x ^{2} + 1 + \frac{1}{x ^{2}}} = \frac{1 - \frac{1}{x ^{2}}}{( x + \frac{1}{x} ) ^{2} - 1}$ . With $t = x + \frac{1}{x}$ , $d t = (1 - \frac{1}{x ^{2}}) d x$ : $\int \frac{d t}{t ^{2} - 1} = \frac{1}{2} lo g \frac{t - 1}{t + 1} + C = \frac{1}{2} lo g \frac{x ^{2} - x + 1}{x ^{2} + x + 1} + C$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the substitution rule.Show answer
$\int f (g (x)) g^{'} (x) d x = \int f (t) d t$ with $t = g (x)$ .
2.What is $\int \frac{f ^{'} ( x )}{f ( x )} d x$ ?Show answer
$lo g ∣ f (x) ∣ + C$ .
3.Which substitution suits $\int \frac{s i n ( l o g x )}{x} d x$ ?Show answer
$t = lo g x$ ; answer $- cos (lo g x) + C$ .
4.What is $\int \frac{d x}{x ^{2} + 9}$ ?Show answer
$\frac{1}{3} tan^{- 1} \frac{x}{3} + C$ .
5.What happens to the limits when you substitute in a definite integral?Show answer
They change to the corresponding values of $t$ .
6.Evaluate $\int x e^{x^{2}} d x$ .Show answer
$\frac{1}{2} e^{x^{2}} + C$ .

Integration by Substitution: frequently asked questions

How do I know what to substitute?

Look for a function inside another function whose derivative also appears as a factor, possibly up to a constant. That inner function is usually the right choice for t.

Do I have to change the limits in a definite integral?

It is the cleanest method: change the limits to t-values and evaluate directly. Alternatively, find the antiderivative, substitute back to x and use the original limits; both give the same answer.

Why do I need + C?

Because many functions have the same derivative, differing by a constant. The indefinite integral represents all of them, so the constant C is part of the answer.

Which NCERT sections cover substitution?

Section 7.3 (Methods of Integration, with 7.3.1 Integration by Substitution) and Section 7.9 (Evaluation of Definite Integrals by Substitution) in Chapter 7 of NCERT Mathematics Part II for Class 12.

Is integration by substitution needed for other methods?

Yes. Partial fractions, special integrals and definite-integral properties often end with a substitution, so it is the most-used method in the chapter.