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MathematicsClass 12Integrals

Integrals | Class 12 Mathematics Notes

By ConceptScroll Team · Published on 17 July 2026 · 3 min read

Integrals – this guide gives you a concise, exam-ready overview of Integrals from Class 12 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

Integration by Substitution

Integration by substitution is a method used to simplify integrals by changing variables. This technique is particularly useful when the integral contains a composite function, i.e., a function inside another function. The method involves substituting a part of the integrand with a new variable t = g(x), which transforms the integral into a simpler form in terms of t. After integrating with respect to t, the original variable x is substituted back. The section explains the step-by-step process: identify the inner function g(x), substitute t = g(x), compute dt = g'(x) dx, rewrite the integral in terms of t and dt, integrate, and then revert to x. Several examples demonstrate how substitution simplifies integrals involving polynomials, trigonometric functions, and exponential functions. The section also discusses the importance of choosing the correct substitution to make the integral manageable.

📊 Diagram: No specific diagrams, but flowcharts showing substitution steps may be included.

🧪 Activity: No specific activity, but exercises involve practicing substitution.

🔗 Connection: Prepares for 'Integration by Parts' which is another technique for integrals involving products.

Frequently asked questions

Find an anti derivative (or integral) of the following functions by the method of inspection. 1. sin 2x 2. cos 3x 3. e2x 4. (ax + b)2 5. sin 2x – 4 e3x

1. ∫ sin 2x dx = -1/2 cos 2x + C

2. ∫ cos 3x dx = 1/3 sin 3x + C

3. ∫ e2x dx = 1/2 e2x + C

4. ∫ (ax + b)^2 dx = ∫ (a^2 x^2 + 2abx + b^2) dx = a^2 ∫ x^2 dx + 2ab ∫ x dx + b^2 ∫ dx = a^2 (x^3/3) + 2ab (x^2/2) + b^2 x + C = (a^2 x^3)/3 + ab x^2 + b^2 x + C

5. ∫ (sin 2x – 4 e3x) dx = ∫ sin 2x dx – 4 ∫ e3x dx = (-1/2) cos 2x – 4 (1/3) e3x + C = (-1/2) cos 2x – (4/3) e3x + C

6. ∫(4 e^{3x} + 1) dx

To evaluate ∫(4 e^{3x} + 1) dx, split the integral:

∫4 e^{3x} dx + ∫1 dx

First integral: ∫4 e^{3x} dx = 4 ∫ e^{3x} dx Using substitution u = 3x, du = 3 dx => dx = du/3 So, ∫ e^{3x} dx = (1/3) e^{3x} + C Therefore, ∫4 e^{3x} dx = 4 * (1/3) e^{3x} = (4/3) e^{3x}

Second integral: ∫1 dx = x + C

Combining: ∫(4 e^{3x} + 1) dx = (4/3) e^{3x} + x + C

7. ∫ x^2 (1 – 1/x^2) dx

Rewrite the integrand:

x^2 (1 – 1/x^2) = x^2 – 1

So, ∫ x^2 (1 – 1/x^2) dx = ∫ (x^2 – 1) dx = ∫ x^2 dx – ∫ 1 dx

Integrate each term: ∫ x^2 dx = (x^3)/3 + C ∫ 1 dx = x + C

Therefore, ∫ x^2 (1 – 1/x^2) dx = (x^3)/3 – x + C

8. ∫ (a x^2 + b x + c) dx

Integrate each term separately:

∫ a x^2 dx = a ∫ x^2 dx = a (x^3 / 3) + C ∫ b x dx = b ∫ x dx = b (x^2 / 2) + C ∫ c dx = c x + C

Therefore, ∫ (a x^2 + b x + c) dx = (a x^3)/3 + (b x^2)/2 + c x + C

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