What is Integration by Parts in Class 12 Maths?

Integration by Parts is part of Chapter 7 (Integrals) in NCERT Class 12 Mathematics. This topic is fundamental for CBSE board exams and competitive tests.

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Integration by Parts

🎓 Class 12 Mathematics CBSE Theory Ch 7 — Integrals ⏱ ~25 min

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7.6 Integration by Parts

In this section, we describe one more method of integration, known as integration by parts^?, that is found quite useful in integrating products of functions.

If \(u\) and \(v\) are any two differentiable functions of a single variable \(x\), then by the product rule of differentiation, we have:

\[\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\]

Integrating both sides, we get:

\[uv = \int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx\]

Integration by Parts Formula

Let \(u = f(x)\) and \(v = g(x)\). Then: \[\int f(x)\cdot g(x)\,dx = f(x)\int g(x)\,dx - \int\left[\frac{d}{dx}f(x)\cdot\int g(x)\,dx\right]dx\]

In words: "Integral of the product of two functions = first function × integral of the second function − integral of [differential coefficient of the first function × integral of the second function]"

ILATE Rule for Choosing First Function

The proper choice of the first function and the second function is significant. Use the ILATE rule — the function that comes first in the following list should be taken as the first function:

I — Inverse trigonometric functions (\(\sin^{-1}x, \tan^{-1}x, \ldots\))
L — Logarithmic functions (\(\log x\))
A — Algebraic functions (\(x, x^2, \ldots\))
T — Trigonometric functions (\(\sin x, \cos x, \ldots\))
E — Exponential functions (\(e^x, a^x, \ldots\))

Remarks

(i) Integration by parts is not applicable to the product of all functions. For instance, the method does not work for \(\int \sqrt{x}\sin x\,dx\), since there is no function whose derivative is \(\sqrt{x}\sin x\).

(ii) While finding the integral of the second function, we do not add any constant of integration. The constant is added only at the final step.

(iii) Usually, if any function is a power of \(x\) or a polynomial in \(x\), we take it as the first function. However, in cases where the other function is an inverse trigonometric function or logarithmic function, we take those as the first function.

Worked Examples

Example 17

Find \(\int x\cos x\,dx\)

Solution

Put \(f(x) = x\) (first function) and \(g(x) = \cos x\) (second function).
Then, integration by parts gives: \[\int x\cos x\,dx = x\int\cos x\,dx - \int\left[\frac{d}{dx}(x)\cdot\int\cos x\,dx\right]dx\] \[= x\sin x - \int 1\cdot\sin x\,dx = x\sin x - (-\cos x) + C = x\sin x + \cos x + C\]

Note: Suppose we take \(f(x) = \cos x\) and \(g(x) = x\). Then \(\int x\cos x\,dx = \cos x\cdot\frac{x^2}{2} - \int\left[-\sin x\cdot\frac{x^2}{2}\right]dx\). This leads to an integral having a higher power of \(x\), which is more complicated. Thus, the proper choice of the first function is crucial.

Example 18

Find \(\int \log x\,dx\)

Solution

To start with, we are unable to guess a function whose derivative is \(\log x\). We take \(\log x\) as the first function and the constant function 1 as the second function. Then: \[\int \log x\cdot 1\,dx = \log x\int 1\,dx - \int\left[\frac{d}{dx}(\log x)\cdot\int 1\,dx\right]dx\] \[= x\log x - \int \frac{1}{x}\cdot x\,dx = x\log x - \int 1\,dx = x\log x - x + C\]

Example 19

Find \(\int e^x\sin x\,dx\)

Solution

Take first function as \(\sin x\) and second function as \(e^x\). \[I = \int e^x\sin x\,dx = \sin x\cdot e^x - \int \cos x\cdot e^x\,dx\] Apply integration by parts again to \(\int e^x\cos x\,dx\): \[= \sin x\cdot e^x - \left[\cos x\cdot e^x - \int(-\sin x)\cdot e^x\,dx\right]\] \[= e^x\sin x - e^x\cos x - \int e^x\sin x\,dx\] \[I = e^x\sin x - e^x\cos x - I\] \[2I = e^x(\sin x - \cos x)\] \[I = \frac{e^x}{2}(\sin x - \cos x) + C\]

Example 20

Find \(\int x\sin^{-1}x\,dx\)

Solution

Take first function as \(\sin^{-1}x\) (by ILATE) and second function as \(x\). \[\int x\sin^{-1}x\,dx = \sin^{-1}x\cdot\frac{x^2}{2} - \int \frac{1}{\sqrt{1-x^2}}\cdot\frac{x^2}{2}\,dx\] \[= \frac{x^2}{2}\sin^{-1}x - \frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}\,dx\] For \(\int \frac{x^2}{\sqrt{1-x^2}}\,dx\): put \(x = \sin\theta\), \(dx = \cos\theta\,d\theta\): \[\int \frac{\sin^2\theta}{\cos\theta}\cdot\cos\theta\,d\theta = \int \sin^2\theta\,d\theta = \frac{\theta}{2} - \frac{\sin 2\theta}{4} + C_1 = \frac{\sin^{-1}x}{2} - \frac{x\sqrt{1-x^2}}{2} + C_1\] Therefore: \[\int x\sin^{-1}x\,dx = \frac{x^2}{2}\sin^{-1}x - \frac{1}{2}\left[\frac{\sin^{-1}x}{2} - \frac{x\sqrt{1-x^2}}{2}\right] + C\] \[= \frac{(2x^2 - 1)}{4}\sin^{-1}x + \frac{x\sqrt{1-x^2}}{4} + C\]

7.6.1 A Special Type: \(\int e^x[f(x) + f'(x)]\,dx\)

Important Result

\[\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C\]

Proof: \(\int e^x[f(x) + f'(x)]\,dx = \int e^x f(x)\,dx + \int e^x f'(x)\,dx\).
Applying integration by parts to the first integral (taking \(e^x\) as second function): \(= e^x f(x) - \int e^x f'(x)\,dx + \int e^x f'(x)\,dx + C = e^x f(x) + C\).

Example: \(\int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right)\,dx\)

Solution

Here \(f(x) = \frac{1}{x}\), so \(f'(x) = -\frac{1}{x^2}\). We have \(f(x) + f'(x) = \frac{1}{x} - \frac{1}{x^2}\).
By the special formula: \(\int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right)\,dx = \frac{e^x}{x} + C\).

7.6.2 Integrals of Some Special Types

Standard Results

(i) \(\int \sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log|x + \sqrt{x^2 - a^2}| + C\)

(ii) \(\int \sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log|x + \sqrt{x^2 + a^2}| + C\)

(iii) \(\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C\)

Example 21

Find \(\int \sqrt{x^2 + 2x + 5}\,dx\)

Solution

Complete the square: \(x^2 + 2x + 5 = (x+1)^2 + 4 = (x+1)^2 + 2^2\).
Put \(x + 1 = y\), so \(dx = dy\): \[\int \sqrt{y^2 + 2^2}\,dy = \frac{y}{2}\sqrt{y^2 + 4} + \frac{4}{2}\log|y + \sqrt{y^2 + 4}| + C\] \[= \frac{(x+1)}{2}\sqrt{x^2 + 2x + 5} + 2\log|(x+1) + \sqrt{x^2 + 2x + 5}| + C\]

Interactive: ILATE Order Explorer

Select a product of two functions and see which one should be the first function

Choose integral:

Choose a product and click Analyse to see the ILATE classification.

Exercise 7.6

Integrate the functions in Exercises 1 to 9:

1. \(\int x\sin x\,dx\)

First fn: \(x\) (A), Second fn: \(\sin x\) (T). By IBP: \(x(-\cos x) - \int 1\cdot(-\cos x)\,dx = -x\cos x + \sin x + C\).

2. \(\int x\sin 3x\,dx\)

\(= x\left(-\frac{\cos 3x}{3}\right) - \int \left(-\frac{\cos 3x}{3}\right)\,dx = -\frac{x\cos 3x}{3} + \frac{\sin 3x}{9} + C\).

3. \(\int x^2 e^x\,dx\)

Apply IBP twice. First: \(x^2 e^x - 2\int x e^x\,dx\). Second: \(\int x e^x\,dx = x e^x - e^x\).
Result: \(x^2 e^x - 2(xe^x - e^x) + C = e^x(x^2 - 2x + 2) + C\).

4. \(\int x\log x\,dx\)

First fn: \(\log x\) (L), Second fn: \(x\) (A). \(\log x\cdot\frac{x^2}{2} - \int \frac{1}{x}\cdot\frac{x^2}{2}\,dx = \frac{x^2}{2}\log x - \frac{x^2}{4} + C\).

5. \(\int x\log 2x\,dx\)

First fn: \(\log 2x\), Second fn: \(x\). \(\frac{x^2}{2}\log 2x - \int \frac{1}{x}\cdot\frac{x^2}{2}\,dx = \frac{x^2}{2}\log 2x - \frac{x^2}{4} + C\).

6. \(\int x^2\log x\,dx\)

\(\log x\cdot\frac{x^3}{3} - \int \frac{1}{x}\cdot\frac{x^3}{3}\,dx = \frac{x^3}{3}\log x - \frac{x^3}{9} + C\).

7. \(\int x\sin^{-1}x\,dx\)

Already solved in Example 20: \(\frac{(2x^2 - 1)}{4}\sin^{-1}x + \frac{x\sqrt{1-x^2}}{4} + C\).

8. \(\int x\tan^{-1}x\,dx\)

First fn: \(\tan^{-1}x\) (I), Second fn: \(x\) (A). \[\frac{x^2}{2}\tan^{-1}x - \frac{1}{2}\int \frac{x^2}{1+x^2}\,dx = \frac{x^2}{2}\tan^{-1}x - \frac{1}{2}\int\left(1 - \frac{1}{1+x^2}\right)\,dx\] \[= \frac{x^2}{2}\tan^{-1}x - \frac{1}{2}(x - \tan^{-1}x) + C = \frac{(x^2+1)}{2}\tan^{-1}x - \frac{x}{2} + C\]

9. \(\int x\cos^{-1}x\,dx\)

First fn: \(\cos^{-1}x\), Second fn: \(x\). \(\frac{x^2}{2}\cos^{-1}x + \frac{1}{2}\int \frac{x^2}{\sqrt{1-x^2}}\,dx\). Using the result from Example 20: \[= \frac{x^2}{2}\cos^{-1}x + \frac{1}{4}\left[\sin^{-1}x - x\sqrt{1-x^2}\right] + C = \frac{(2x^2 - 1)}{4}\cos^{-1}x - \frac{x\sqrt{1-x^2}}{4} + C\]

Choose the correct answer in Exercises 10 and 11:

10. \(\int \sqrt{x^2 + 1}\,dx\) is equal to:
(A) \(\frac{x}{2}\sqrt{x^2+1} + \frac{1}{2}\log|x+\sqrt{x^2+1}| + C\)
(B) \(\frac{2}{3}(x^2+1)^{3/2} + C\)
(C) \(\frac{2}{3}x(x^2+1)^{3/2} + C\)
(D) \(\frac{x^2}{2}\sqrt{x^2+1} + \frac{1}{2}x^2\log|x+\sqrt{x^2+1}| + C\)

Using formula (ii) with \(a = 1\): \(\frac{x}{2}\sqrt{x^2+1} + \frac{1}{2}\log|x + \sqrt{x^2+1}| + C\). Answer: (A)

11. \(\int \sqrt{x^2 - 8x + 7}\,dx\) is equal to:

Complete the square: \(x^2 - 8x + 7 = (x-4)^2 - 9\). Let \(y = x - 4\): \[\int \sqrt{y^2 - 3^2}\,dy = \frac{y}{2}\sqrt{y^2 - 9} - \frac{9}{2}\log|y + \sqrt{y^2 - 9}| + C\] \[= \frac{(x-4)}{2}\sqrt{x^2-8x+7} - \frac{9}{2}\log|(x-4)+\sqrt{x^2-8x+7}| + C\]

Activity: ILATE Classification Race

L4 Analyse

Challenge: For each integral, identify the first function (using ILATE) and the second function in under 10 seconds.

Write these on cards: (a) \(\int x^2 \sin x\,dx\) (b) \(\int e^x \log x\,dx\) (c) \(\int \tan^{-1}x\,dx\) (d) \(\int x^3 e^{2x}\,dx\)
For each, circle the first function based on ILATE order.
For (c), note: the second function is 1 (the constant function), since \(\log x\) and inverse trig functions have no easy integral on their own.

(a) First: \(x^2\) (A), Second: \(\sin x\) (T).

(b) First: \(\log x\) (L), Second: \(e^x\) (E).

(d) First: \(x^3\) (A), Second: \(e^{2x}\) (E).

Competency-Based Questions

Scenario: In a physics experiment, the displacement of a vibrating string is modelled as \(y(t) = \int_0^t \tau e^{-\tau}\sin\tau\,d\tau\). A student needs to first evaluate the indefinite integral \(\int x e^{-x}\sin x\,dx\) to find the displacement at any time.

Q1. In the integral \(\int x e^{-x}\,dx\), using ILATE the first function should be:

L1 Remember

(a) \(e^{-x}\) because E comes before A
(b) \(x\) because A comes before E in ILATE
(c) Either one works equally well
(d) The ILATE rule does not apply here

Answer: (b). In ILATE, Algebraic (A) comes before Exponential (E). So \(x\) is the first function, \(e^{-x}\) the second.

Q2. Evaluate \(\int x e^{-x}\,dx\) step by step.

L3 Apply

Answer: First fn: \(x\), Second fn: \(e^{-x}\). \(\int e^{-x}\,dx = -e^{-x}\).
\(\int x e^{-x}\,dx = x(-e^{-x}) - \int 1\cdot(-e^{-x})\,dx = -xe^{-x} + \int e^{-x}\,dx = -xe^{-x} - e^{-x} + C = -e^{-x}(x+1) + C\).

Q3. Apply the special formula to evaluate \(\int e^x\left(\sin^{-1}x + \frac{1}{\sqrt{1-x^2}}\right)\,dx\). Explain why this formula applies.

L4 Analyse

Answer: Let \(f(x) = \sin^{-1}x\). Then \(f'(x) = \frac{1}{\sqrt{1-x^2}}\). The integrand is \(e^x[f(x) + f'(x)]\). By the special formula: \(\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C = e^x \sin^{-1}x + C\). The formula applies because the integrand is precisely \(e^x\) times the sum of a function and its derivative.

Q4. A student claims that \(\int \log x\,dx\) cannot be solved by integration by parts because there is only one function. Evaluate this claim and correct any misconception.

L5 Evaluate

Answer: The claim is incorrect. We can always write any function as a product with 1: \(\int \log x\,dx = \int \log x \cdot 1\,dx\). Take \(\log x\) as the first function and 1 as the second. IBP gives: \(x\log x - x + C\). This technique works for any function that is hard to integrate directly but easy to differentiate — including \(\sin^{-1}x\), \(\tan^{-1}x\), etc.

Assertion–Reason Questions

Assertion (A): \(\int e^x(\sin x + \cos x)\,dx = e^x\sin x + C\).
Reason (R): \(\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C\).

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Both true. Here \(f(x) = \sin x\), \(f'(x) = \cos x\). The integrand is \(e^x[f + f']\), so the result is \(e^x f(x) = e^x\sin x + C\). R is the general formula that directly gives A.

Assertion (A): In \(\int x^2 \cos x\,dx\), we choose \(x^2\) as the first function.
Reason (R): According to ILATE, algebraic functions are chosen before trigonometric functions.

(a) Both true, R explains A.

(b) Both true, R doesn't explain A.

(c) A true, R false.

(d) A false, R true.

Answer: (a) — Both true. ILATE order is I-L-A-T-E. Since A (Algebraic) comes before T (Trigonometric), \(x^2\) is chosen as the first function. R correctly states the ILATE rule and explains the choice in A.

Frequently Asked Questions

What is the formula for integration by parts?

Integral of u*v dx = u * integral(v dx) - integral of [du/dx * integral(v dx)] dx. Derived from the product rule of differentiation, used for products of two functions.

What is the ILATE rule?

ILATE prioritizes choosing the first function: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. The function appearing first in this order is chosen as u.

When should you use integration by parts?

When the integrand is a product of two different function types like x*sin(x), x*e^x, or log(x). Also for integrals of log(x) and inverse trig functions alone.

What is the special e^x formula?

Integral of e^x[f(x) + f'(x)]dx = e^x * f(x) + C. This shortcut applies when e^x multiplies the sum of a function and its derivative.

How to handle repeated integration by parts?

If the original integral reappears after two applications, collect integral terms on one side and solve algebraically. Common with e^x*sin(x) or e^x*cos(x).

Frequently Asked Questions — Integrals

What is Integration by Parts in NCERT Class 12 Mathematics?

Integration by Parts is a key concept covered in NCERT Class 12 Mathematics, Chapter 7: Integrals. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Integration by Parts step by step?

To solve problems on Integration by Parts, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 7: Integrals?

The essential formulas of Chapter 7 (Integrals) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Integration by Parts important for the Class 12 board exam?

Integration by Parts is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Integration by Parts?

Common mistakes in Integration by Parts include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Integration by Parts?

End-of-chapter NCERT exercises for Integration by Parts cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 7, and solve at least one previous-year board paper to consolidate your understanding.

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