Chapter 7: Integrals - Board Exam Notes

Exam Weightage & Blueprint

Total: ~10-12 Marks

Integrals is the heavyweight champion of Class 12 Maths. It forms the basis for Application of Integrals and Differential Equations. Mastery of integration techniques is non-negotiable.

Question Type	Marks	Frequency	Focus Topic
MCQ	1	Very High	Standard Formulas, Simple Definite Integrals
Short Answer (2M)	2	High	Substitution Method, Trigonometric Identities
Short Answer (3M)	3	High	Partial Fractions, Integration by Parts, Special Integrals
Long Answer	5	Very High	Properties of Definite Integrals (especially P4)

Last 24-Hour Checklist

Standard Formulas: $\int x^n, \int e^x, \int \frac{1}{x}, \int \sin x$
Substitution: Identify $f(x)$ and $f'(x)$ in the integrand
Special Integrals: $\frac{1}{x^2 \pm a^2}, \frac{1}{\sqrt{a^2 \pm x^2}}$
By Parts (ILATE): $\int uv = u\int v - \int(u'\int v)$

Partial Fractions: $\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$
Cancellation Form: $\int e^x[f(x) + f'(x)]dx = e^xf(x)$
Property P4: $\int_0^a f(x)dx = \int_0^a f(a-x)dx$
Even/Odd Property: $\int_{-a}^a f(x)dx$

Indefinite Integrals: Key Methods

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

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

3. Special Integrals (Memorize!)

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

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

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

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

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

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

Definite Integrals & Properties

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

Important Properties

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):
$\int_{-a}^a f(x) dx = \begin{cases} 2\int_0^a f(x) dx, & \text{if } f(-x) = f(x) \text{ (Even)} \\ 0, & \text{if } f(-x) = -f(x) \text{ (Odd)} \end{cases}$

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

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

Q2. Evaluate $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$. (4 Marks)

Step 1: Apply Property P4 1 Mark

Let $I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx \quad \dots (1)$

$I = \int_0^{\pi/2} \frac{\sqrt{\sin(\pi/2 - x)}}{\sqrt{\sin(\pi/2 - x)} + \sqrt{\cos(\pi/2 - x)}} dx$

$I = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx \quad \dots (2)$

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

$2I = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$

$2I = \int_0^{\pi/2} 1 \, dx = [x]_0^{\pi/2} = \frac{\pi}{2}$.

Step 3: Final Answer 1 Mark

$I = \frac{\pi}{4}$.

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

2022 (3 Marks): Evaluate $\int_0^{\pi/2} \log(\tan x) dx$.
Ans: Let $I = \int_0^{\pi/2} \log(\tan x) dx$. Using P4, $I = \int_0^{\pi/2} \log(\tan(\pi/2-x)) dx = \int_0^{\pi/2} \log(\cot x) dx$.
Adding both: $2I = \int_0^{\pi/2} (\log \tan x + \log \cot x) dx = \int_0^{\pi/2} \log(\tan x \cdot \cot x) dx = \int_0^{\pi/2} \log(1) dx = 0$.
So $I = 0$.

2020 (3 Marks): Find $\int \frac{x^2+1}{x^2-5x+6} dx$.
Hint: Degree of numerator $\ge$ 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 $x=a\sin\theta$. If $a^2+x^2$, try $x=a\tan\theta$.

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

Tip 3: If the integrand is a rational function with denominator degree $\ge$ 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: $[\tan^{-1}x]_1^{\sqrt{3}} = \tan^{-1}\sqrt{3} - \tan^{-1}1 = \frac{\pi}{3} - \frac{\pi}{4} = \frac{\pi}{12}$.

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 ($u=x, v=\sin x$). $-x\cos x - \int (-\cos x) dx = -x\cos x + \sin x + C$.

Level 3: Advanced (5 Marks Each)

Q5. Evaluate $\int_0^{\pi} \frac{x \sin x}{1+\cos^2 x} dx$.

Hint: Use P4 ($x \to \pi-x$). $I = \int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2 x} dx$.
Add original and new: $2I = \pi \int_0^\pi \frac{\sin x}{1+\cos^2 x} dx$. Put $\cos x = t$.
Ans: $\frac{\pi^2}{4}$.