What is the weightage of Chapter 7 Integrals in CBSE Class 12?

Integrals is among the highest-yield chapters in Class 12 Maths, contributing roughly 8 to 10 marks in Unit III across Section B, C, and D.

What is the priority rule for choosing the first function in Integration by Parts?

Use the ILATE rule in order: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, and Exponential functions.

What is the King Property of Definite Integrals?

Integral from a to b of f(x) dx = Integral from a to b of f(a + b - x) dx. For limits from 0 to a, it states: Integral from 0 to a of f(x) dx = Integral from 0 to a of f(a - x) dx.

How do you integrate an odd function over symmetric limits [-a, a]?

If f(-x) = -f(x), the function is odd and the integral from -a to a of f(x) dx evaluates directly to 0.

Class 12 Maths Chapter 7: Integrals Notes & Formulas

2026?27 Syllabus Focus

The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.

Integration by substitution, partial fractions and parts.
The standard integral forms listed in the CBSE syllabus.
The Fundamental Theorem of Calculus (without proof), properties of definite integrals and evaluation.

CBSE Mathematics (Code 041) curriculum, 2026?27

Last 24-Hour Checklist

Indefinite Integrals: Key Methods

Integration as Inverse of Differentiation:
If $F'(x)=f(x)$, then $$\int f(x)\,dx = F(x)+C$$ where $C$ is an arbitrary constant.

Basic Properties of Indefinite Integrals:

$\int [f(x)\pm g(x)]dx = \int f(x)dx \pm \int g(x)dx$

$\int kf(x)\,dx = k\int f(x)\,dx \quad (k constant)$

$\int 0\,dx = C$

1. Integration by Substitution

If integral is of the form $\int f(g(x)) g'(x) dx$, put $g(x) = t \Rightarrow g'(x)dx = dt$.
Example: $\int 2x \sin(x^2+1) dx$. Put $x^2+1=t \Rightarrow 2x dx = dt$.

2. Integration by Parts

$$\int u v \, dx = u \int v \, dx - \int ( \frac{du}{dx} \int v \, dx ) dx$$ Order of selection for $u$ (ILATE):
Inverse Trig, Logarithmic, Algebraic, Trigonometric, Exponential.

Special Form (Cancellation Law):

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

Example: $\int e^x (\sin x + \cos x) dx = e^x \sin x + C$.

3. Integration using Trigonometric Identities

Before integrating trig powers/products, simplify using identities:

$\sin^2 x=\frac{1-\cos 2x}{2},\quad \cos^2 x=\frac{1+\cos 2x}{2}$

$\sin x\cos x=\frac{\sin 2x}{2},\quad 1+\tan^2x=\sec^2x,\quad 1+\cot^2x=\csc^2x$

Example pattern: $∫ \sin^2x\,dx,\ ∫ \sin x\cos x\,dx,\ ∫ \frac{dx}{1+\tan^2x}$

4. Integration by Partial Fractions

Case A: Distinct Linear Factors

$\frac{P(x)}{(x-a)(x-b)}=\frac{A}{x-a}+\frac{B}{x-b}$

Case B: Repeated Linear Factor

$\frac{P(x)}{(x-a)^2}=\frac{A}{x-a}+\frac{B}{(x-a)^2}$

Case C: Irreducible Quadratic Factor

$\frac{P(x)}{(x^2+px+q)}=\frac{Ax+B}{x^2+px+q}$

Important

If degree(numerator) $≥$ degree(denominator), divide first, then decompose.

5. Special Integrals (Memorize!)

$\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \log|\frac{x-a}{x+a}| + C$

$\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log|\frac{a+x}{a-x}| + C$

$\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1}\frac{x}{a} + C$

$\int \frac{dx}{\sqrt{x^2 - a^2}} = \log|x + \sqrt{x^2 - a^2}| + C$

$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C$

$\int \frac{dx}{\sqrt{x^2 + a^2}} = \log|x + \sqrt{x^2 + a^2}| + C$

6. Some Other Standard Forms

$\int \frac{dx}{ax+b}=\frac{1}{a}\log|ax+b|+C$

$\int \tan x\,dx=\log|\sec x|+C$

$\int \cot x\,dx=\log|\sin x|+C$

$\int \sec x\,dx=\log|\sec x+\tan x|+C$

$\int \csc x\,dx=\log|\csc x-\cot x|+C$

$\int \sec x\tan x\,dx=\sec x + C,\ \int \csc x\cot x\,dx=-\csc x + C$

Definite Integrals & Properties

Fundamental Theorem of Calculus:
$\int _a^b f(x) dx = F(b) - F(a)$, where $F'(x) = f(x)$.

Important Properties

$\int _a^a f(x)\,dx = 0$

$\int _a^b f(x)\,dx = -\int _b^a f(x)\,dx$

$\int _a^b f(x)\,dx + \int _b^c f(x)\,dx = \int _a^c f(x)\,dx$

$\int _a^b [kf(x)\pm g(x)]\,dx = k\int _a^b f(x)\,dx \pm \int _a^b g(x)\,dx$

Property P4 (King's Property)

$\int _0^a f(x) dx = \int _0^a f(a-x) dx$

Use when: Denominator remains same after replacing $x$ with $a-x$.

Property P3

$\int _a^b f(x) dx = \int _a^b f(a+b-x) dx$

(General version of P4)

Property P7 (Even/Odd Functions):
$∫_-a^a f(x) dx = \begin{cases} 2∫_0^a f(x) dx, & if f(-x) = f(x) (Even Function) \\ 0, & if f(-x) = -f(x) (Odd Function) \end{cases}$

Quick Trick: Any odd power of sine or $x$ (e.g., $x^3, \sin^5 x$) integrated from $-a$ to $a$ is $0$.

Special Consequence of King's Property:
$\int _0^{π/2} f(\sin x,\cos x)\,dx = \int _0^{π/2} f(\cos x,\sin x)\,dx$
Very common in board exams for integrals like $∫_0^π/2 \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \frac{π}{4}$.

Useful Symmetry Results:

If $f(x)=f(a-x)$, then $\int _0^a f(x)\,dx = 2\int _0^{a/2} f(x)\,dx$.

If $f(x)+f(a-x)=k$, then $\int _0^a f(x)\,dx = \frac{ka}{2}$.

Definite Integral Substitution Rule:
When substituting $x=g(t)$ in $\int _a^b f(x)\,dx$, always change limits first, then integrate in the new variable.

⚡ High-Yield Integral Formulas & Master Properties (CBSE 4-6 Mark Questions):

The King Property $P_4$ (Solves 80% of Board Definite Integrals):
$$\int_0^a f(x)dx = \int_0^a f(a - x)dx \quad \text{and} \quad \int_a^b f(x)dx = \int_a^b f(a + b - x)dx$$ Standard Strategy: Let $I = \int_a^b f(x)dx$. Apply property to get $I = \int_a^b f(a+b-x)dx$. Add the two equations: $2I = \int_a^b [f(x) + f(a+b-x)]dx$.
Even / Odd Function Integral Property ($P_7$):
$$\int_{-a}^a f(x)dx = \begin{cases} 2 \int_0^a f(x)dx & \text{if } f(-x) = f(x) \text{ (Even Function)} \\ 0 & \text{if } f(-x) = -f(x) \text{ (Odd Function)} \end{cases}$$ • MCQ Shortcut: $\int_{-\pi/2}^{\pi/2} \sin^7 x \, dx = 0$ (since $\sin^7(-x) = -\sin^7 x$).
The Special Exponential-Product Pattern:
$$\int e^x [f(x) + f'(x)]dx = e^x f(x) + C$$ • Example: $\int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right)dx = e^x \cdot \frac{1}{x} + C$.
The 3 Radical Standard Integrals:
• $\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$
• $\int \sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log|x + \sqrt{x^2 + a^2}| + C$
• $\int \sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log|x + \sqrt{x^2 - a^2}| + C$
Improper Rational Fraction Trap:
For $\int \frac{P(x)}{Q(x)}dx$, if $\deg(P) \ge \deg(Q)$, you MUST perform polynomial long division first: $\frac{P(x)}{Q(x)} = T(x) + \frac{R(x)}{Q(x)}$ before applying partial fractions.
📌 Syllabus Clarification: Evaluation of definite integrals as the limit of a sum (NCERT Ex 7.8) is removed from the active CBSE syllabus.

Solved Examples (Board Marking Scheme)

Q1. Evaluate $\int \frac{x}{(x+1)(x+2)} dx$. (3 Marks)

Step 1: Partial Fractions 1 Mark

Let $\frac{x}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$.

$x = A(x+2) + B(x+1)$.

Put $x = -1 \Rightarrow -1 = A(1) \Rightarrow A = -1$.

Put $x = -2 \Rightarrow -2 = B(-1) \Rightarrow B = 2$.

Step 2: Integrate 1 Mark

$\int (\frac{-1}{x+1} + \frac{2}{x+2}) dx = -\log|x+1| + 2\log|x+2| + C$.

Step 3: Simplify (Optional) 1 Mark

$= \log|x+2|^2 - \log|x+1| + C = \log|\frac{(x+2)^2}{x+1}| + C$.

Q2. Evaluate $\int _0^{π/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$. (4 Marks)

Step 1: Apply Property P4 1 Mark

Let $I = ∫_0^π/2 \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \quad \dots (1)$

$I = ∫_0^π/2 \frac{\sqrt{\sin(π/2 - x)}}{\sqrt{\sin(π/2 - x)} + \sqrt{\cos(π/2 - x)}} dx$

$I = \int _0^{π/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \quad \dots (2)$

Step 2: Add (1) and (2) 2 Marks

$2I = \int _0^{π/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$

$2I = \int _0^{π/2} 1 \, dx = [x]_0^{π/2} = \frac{π}{2}$.

Step 3: Final Answer 1 Mark

$I = \frac{π}{4}$.

Q3. Find $\int e^x (\tan^{-1}x + \frac{1}{1+x^2}) dx$. (2 Marks)

Solution: 2 Marks

We know that $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$.

Here, let $f(x) = \tan^{-1}x$, then $f'(x) = \frac{1}{1+x^2}$.

So, Integral $= e^x \tan^{-1}x + C$.

Previous Year Questions (PYQs)

2023 (2 Marks): Find $\int \frac{1}{\sqrt{x^2+2x+2}} dx$.
Ans: Complete the square: $x^2+2x+2 = (x+1)^2 + 1^2$.
Integral becomes $∫ \frac{dx}{\sqrt{(x+1)^2 + 1^2}} = \log| (x+1) + \sqrt{(x+1)^2+1} | + C = \log|x+1+\sqrt{x^2+2x+2}| + C$.

2022 (3 Marks): Evaluate $\int _0^{π/2} \log(\tan x) dx$.
Ans: Let $I = \int _0^{π/2} \log(\tan x) dx$. Using P4, $I = ∫_0^π/2 \log(\tan(π/2-x)) dx = ∫_0^π/2 \log(\cot x) dx$.
Adding both: $2I = ∫_0^π/2 (\log \tan x + \log \cot x) dx = ∫_0^π/2 \log(\tan x · \cot x) dx = ∫_0^π/2 \log(1) dx = 0$.
So $I = 0$.

2020 (3 Marks): Find $\int \frac{x^2+1}{x^2-5x+6} dx$.
Hint: Degree of numerator $≥$ denominator. Divide first. $x^2+1 = 1(x^2-5x+6) + (5x-5)$.
Integral $= \int 1 dx + \int \frac{5x-5}{(x-2)(x-3)} dx$. Use partial fractions for the second part.

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: Forgetting the constant of integration $+C$ in indefinite integrals. This costs 0.5 to 1 mark.

Mistake 2: In substitution for definite integrals, forgetting to change the limits (lower and upper bounds).

Mistake 3: Wrong order in Integration by Parts. Remember ILATE. If you pick wrong $u$, the integral becomes harder.

Scoring Tips

Tip 1: If you see $\sqrt{a^2-x^2}$, try $x=a\sinθ$. If $a^2+x^2$, try $x=a\tanθ$.

Tip 2: For definite integrals with limits $0$ to $a$ or $0$ to $π/2$, always check Property P4 first. It solves 80% of such problems.

Tip 3: If the integrand is a rational function with denominator degree $≥$ numerator, ALWAYS divide first.

Practice Problems (Self-Assessment)

Level 1: Basic (1 Mark Each)

Q1. Find $\int (e^{2x} + x^2) dx$.

Answer: $\frac{e^{2x}}{2} + \frac{x^3}{3} + C$.

Q2. Evaluate $\int _1^{\sqrt{3}} \frac{dx}{1+x^2}$.

Answer: $[\tan^{-1}x]_1^\sqrt{3} = \tan^{-1}\sqrt{3} - \tan^{-1}1 = \frac{π}{3} - \frac{π}{4} = \frac{π}{12}$.

Level 2: Intermediate (2-3 Marks Each)

Q3. Find $\int \frac{\sin^3 x + \cos^3 x}{\sin^2 x \cos^2 x} dx$.

Hint: Split terms: $\frac{\sin x}{\cos^2 x} + \frac{\cos x}{\sin^2 x} = \sec x \tan x + \csc x \cot x$.
Ans: $\sec x - \csc x + C$.

Q4. Evaluate $\int x \sin x dx$.

Answer: By parts ($u=x, v=\sin x$). $-x\cos x - ∫ (-\cos x) dx = -x\cos x + \sin x + C$.

Level 3: Advanced (5 Marks Each)

Q5. Evaluate $\int _0^{π} \frac{x \sin x}{1+\cos^2 x} dx$.

Hint: Use P4 ($x \to π-x$). $I = ∫_0^π \frac{(π-x)\sin x}{1+\cos^2 x} dx$.
Add original and new: $2I = π \int _0^π \frac{\sin x}{1+\cos^2 x} dx$. Put $\cos x = t$.
Ans: $\frac{π^2}{4}$.

Chapter 7 Board Mastery Test

Practice test • 10 marks • 25 min

Q1 (1 Mark): Evaluate $\int e^x (\frac{1}{x} - \frac{1}{x^2}) dx$.

Q2 (2 Marks): Evaluate $\int \frac{dx}{x^2 - 16}$.

Q3 (3 Marks): Evaluate $\int \frac{x}{(x-1)(x-2)} dx$ using partial fractions.

Q4 (4 Marks): Evaluate $\int _0^{π/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ using properties of definite integrals.

Marking Scheme & Solutions

Solution 1:

Using standard form $\int e^x [f(x) + f'(x)] dx = e^x f(x) + C$:

Here $f(x) = \frac{1}{x} \implies f'(x) = -\frac{1}{x^2}$.

Therefore, $\int e^x (\frac{1}{x} - \frac{1}{x^2}) dx = \mathbf{\frac{e^x}{x} + C}$. [1 Mark]

Solution 2:

$\int \frac{dx}{x^2 - 4^2} = \frac{1}{2(4)} \log|\frac{x-4}{x+4}| + C = \mathbf{\frac{1}{8}\log|\frac{x-4}{x+4}| + C}$. [2 Marks]

Solution 3:

Let $\frac{x}{(x-1)(x-2)} = \frac{A}{x-1} + \frac{B}{x-2} \implies x = A(x-2) + B(x-1)$.

Put $x = 1 \implies 1 = -A \implies A = -1$. Put $x = 2 \implies 2 = B \implies B = 2$. [1 Mark]

$\int (\frac{-1}{x-1} + \frac{2}{x-2}) dx = -\log|x-1| + 2\log|x-2| + C = \mathbf{\log\frac{(x-2)^2}{|x-1|} + C}$. [2 Marks]

Solution 4:

Let $I = \int _0^{π/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ --- (1)

Using property $\int _0^a f(x)dx = \int _0^a f(a-x)dx$:

$I = \int _0^{π/2} \frac{\sqrt{\sin(\frac{π}{2}-x)}}{\sqrt{\sin(\frac{π}{2}-x)} + \sqrt{\cos(\frac{π}{2}-x)}} dx = \int _0^{π/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ --- (2) [1.5 Marks]

Adding (1) and (2):

$2I = \int _0^{π/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int _0^{π/2} 1 \cdot dx = [x]_0^{π/2} = \frac{π}{2}$. [1.5 Marks]

$2I = \frac{π}{2} \implies \mathbf{I = \frac{π}{4}}$. [1 Mark]

Chapter 7: Integrals

Overview