Integrals — Complete Guide with Solved Examples

What Is Integration, Really?

If differentiation asks “how fast is this changing?”, integration asks “how much has accumulated?” These two questions are mirror images of each other — and that relationship is the entire foundation of calculus.

Think of it this way: if you know a car’s velocity at every instant, integration tells you the total distance covered. If you know how quickly a tank fills, integration gives you total volume. The geometric picture is the area under a curve — but the deeper idea is accumulation.

We have two flavours to work with. A definite integral gives a number — the actual area between $x = a$ and $x = b$ . An indefinite integral gives a family of functions — all possible antiderivatives, differing by a constant.

\int f(x)\, dx = F(x) + C \quad \text{where } F'(x) = f(x)

That $+ C$ is not optional. Every time you write an indefinite integral without it, you’re technically wrong — and board examiners notice.

The reason Class 12 students find integrals harder than derivatives is simple: differentiation has mechanical rules that always work. Integration is more like solving a puzzle — you have to recognise the pattern first, then choose the technique. The good news? JEE Main repeats patterns. Once you’ve seen enough of them, you’ll start recognising the puzzle pieces.

Key Terms and Definitions

Antiderivative (Primitive): A function $F(x)$ such that $F'(x) = f(x)$ . Integration finds antiderivatives.

Indefinite Integral: $\int f(x)\, dx = F(x) + C$ . No limits, gives a general function.

Definite Integral: $\int_a^b f(x)\, dx = F(b) - F(a)$ . Has limits $a$ (lower) and $b$ (upper), gives a number.

Integrand: The function being integrated — the $f(x)$ part.

Variable of Integration: The $dx$ part. It tells us with respect to what we’re integrating. Don’t treat it as decoration — it’s crucial in substitution.

Constant of Integration ( $C$ ): Added to all indefinite integrals because differentiating any constant gives zero. The general solution is a family of curves.

Area Under a Curve: Geometrically, $\int_a^b f(x)\, dx$ is the signed area between the curve $y = f(x)$ and the x-axis from $a$ to $b$ . Area below the x-axis contributes negatively.

Core Methods

Standard Formulas (Non-Negotiable)

Memorise these cold. Every technique eventually reduces to one of these.

\int x^n\, dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1

\int \frac{1}{x}\, dx = \ln|x| + C

\int e^x\, dx = e^x + C

\int a^x\, dx = \frac{a^x}{\ln a} + C

\int \sin x\, dx = -\cos x + C

\int \cos x\, dx = \sin x + C

\int \sec^2 x\, dx = \tan x + C

\int \cosec^2 x\, dx = -\cot x + C

\int \sec x \tan x\, dx = \sec x + C

\int \cosec x \cot x\, dx = -\cosec x + C

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

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

\int \frac{1}{x\sqrt{x^2-1}}\, dx = \sec^{-1} x + C

Method 1: Substitution (U-Substitution)

The most important technique. The idea: if you see a function and its derivative inside the same integral, substitute.

When to use: When the integrand has a composite function like $f(g(x)) \cdot g'(x)$ .

Process:

Identify a substitution $u = g(x)$
Compute $du = g'(x)\, dx$ , so $dx = \frac{du}{g'(x)}$
Rewrite everything in terms of $u$
Integrate, then substitute back

Worked Example: $\int 2x \cdot e^{x^2}\, dx$

Let $u = x^2$ , so $du = 2x\, dx$ .

\int e^u\, du = e^u + C = e^{x^2} + C

The “tell” for substitution: look for a function and something that looks like its derivative sitting together. In $\int \frac{\ln x}{x}\, dx$ , you see $\ln x$ and $\frac{1}{x}$ (derivative of $\ln x$ ) — classic substitution signal.

Method 2: Integration by Parts

Formula:

\int u\, dv = uv - \int v\, du

ILATE rule for choosing $u$ (pick whichever comes first in this order):

I — Inverse trig ( $\sin^{-1}x$ , $\tan^{-1}x$ )
L — Logarithmic ( $\ln x$ , $\log x$ )
A — Algebraic ( $x^2$ , $x^3$ , polynomials)
T — Trigonometric ( $\sin x$ , $\cos x$ )
E — Exponential ( $e^x$ , $2^x$ )

Worked Example: $\int x \cos x\, dx$

By ILATE: $u = x$ (Algebraic), $dv = \cos x\, dx$

$du = dx$ , $v = \sin x$

\int x \cos x\, dx = x \sin x - \int \sin x\, dx = x \sin x + \cos x + C

The formula $\int e^x[f(x) + f'(x)]\, dx = e^x f(x) + C$ appears almost every year in JEE Main. Recognise it on sight — don’t rederive it in the exam.

Method 3: Partial Fractions

Use when the integrand is a rational function (polynomial divided by polynomial) where the degree of numerator < degree of denominator.

Standard decompositions:

\frac{px + q}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}

\frac{px + q}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}

\frac{px^2 + qx + r}{(x-a)(x^2+bx+c)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+bx+c}

Worked Example: $\int \frac{1}{x^2 - 1}\, dx$

\frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}

Solving: $A = \frac{1}{2}$ , $B = -\frac{1}{2}$

\int \frac{1}{x^2-1}\, dx = \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C = \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C

Method 4: Definite Integral Properties

These are high-weightage in both CBSE and JEE. Knowing properties often reduces a 5-step problem to 2 steps.

\int_a^b f(x)\, dx = \int_a^b f(a+b-x)\, dx \quad \text{(King Property)}

\int_0^a f(x)\, dx = \int_0^a f(a-x)\, dx

\int_{-a}^{a} f(x)\, dx = \begin{cases} 2\int_0^a f(x)\, dx & \text{if } f \text{ is even} \\ 0 & \text{if } f \text{ is odd} \end{cases}

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

The King Property is the most powerful. When you see $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\, dx$ , don’t try to integrate directly — apply King.

Solved Examples

Example 1 — CBSE Level

Evaluate: $\int \frac{x^2 + 1}{x^2 - 1}\, dx$

Degree of numerator = degree of denominator, so first do polynomial division:

\frac{x^2+1}{x^2-1} = 1 + \frac{2}{x^2-1}

Now apply partial fractions to $\frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1}$

\int \left(1 + \frac{1}{x-1} - \frac{1}{x+1}\right) dx = x + \ln|x-1| - \ln|x+1| + C

Example 2 — JEE Main Level

Evaluate: $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\, dx$

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

By King Property (replace $x$ with $\frac{\pi}{2} - x$ ):

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

Adding both expressions:

2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\, dx = \int_0^{\pi/2} 1\, dx = \frac{\pi}{2}

I = \frac{\pi}{4}

This exact question appeared in JEE Main 2023 Session 1. The trap is trying to simplify the integrand directly — always check if a property applies first.

Example 3 — JEE Advanced Level

Evaluate: $\int \sqrt{\tan x}\, dx$

Let $t = \sqrt{\tan x}$ , so $t^2 = \tan x$ , $2t\, dt = \sec^2 x\, dx = (1 + t^4)\, dx$

\int \sqrt{\tan x}\, dx = \int \frac{2t^2}{1+t^4}\, dt

Split: $\frac{2t^2}{1+t^4} = \frac{(t^2+1) + (t^2-1)}{1+t^4}$

Divide numerator and denominator by $t^2$ :

\int \frac{1 + 1/t^2}{t^2 + 1/t^2}\, dt + \int \frac{1 - 1/t^2}{t^2 + 1/t^2}\, dt

Each becomes a standard form by completing the square. This problem tests whether you can reduce to $\int \frac{dt}{u^2 \pm k^2}$ form.

Exam-Specific Tips

CBSE Class 12: Integration carries about 20 marks across the paper (integrals + applications). The 3-mark questions almost always test substitution or by-parts. The 5-mark question typically combines a method with evaluation of a definite integral. Writing intermediate steps matters — partial credit is real.

JEE Main: Expect 2-3 questions from integrals per session. High-frequency topics: definite integral using properties (King property especially), integrals of the form $\int e^x[f(x) + f'(x)]\, dx$ , and integrals reducible to $\tan^{-1}$ or $\ln$ form. Time per question: roughly 2 minutes. If a question isn’t yielding in 90 seconds, skip and return.

JEE Advanced: Integration appears in longer, multi-part problems. Expect integration combined with differential equations, or definite integrals where you need to estimate bounds. The $\sqrt{\tan x}$ family, Walli’s formula applications, and bounded area problems are recurring themes. Walli’s formula ( $\int_0^{\pi/2} \sin^n x\, dx$ ) saves significant time in Paper 2.

Common Mistakes to Avoid

Dropping the constant $C$ in indefinite integrals. In CBSE, this costs you marks on every single indefinite integral. Make it automatic.

Power rule on $x^{-1}$ . Students write $\int x^{-1}\, dx = \frac{x^0}{0} + C$ . Division by zero. The correct answer is $\ln|x| + C$ — this is a separate formula to memorise, not a special case of the power rule.

Forgetting the modulus in $\ln|x|$ . When the argument of a logarithm comes from integration, it needs the modulus sign because $\ln$ is only defined for positive numbers, but $x$ could be negative. Write $\ln|x-1|$ , not $\ln(x-1)$ .

Wrong ILATE order. Students often pick the exponential as $u$ when there’s a polynomial present. By ILATE, Algebraic beats Exponential — so in $\int x^2 e^x\, dx$ , let $u = x^2$ , not $e^x$ .

Signed area vs. actual area. $\int_{-1}^{1} x^3\, dx = 0$ because $x^3$ is odd. But the area enclosed between the curve and x-axis is not zero — it’s 2 times the area from 0 to 1. For “find the area” problems, integrate $|f(x)|$ , not $f(x)$ directly. Split the integral at zeros of $f$ .

Practice Questions

Q1. Evaluate $\int \frac{1}{\sqrt{x} + x}\, dx$

Let $\sqrt{x} = t$ , so $x = t^2$ , $dx = 2t\, dt$ .

\int \frac{2t\, dt}{t + t^2} = \int \frac{2}{1+t}\, dt = 2\ln|1 + \sqrt{x}| + C

Q2. Evaluate $\int_0^1 x(1-x)^9\, dx$

By King Property, $\int_0^1 x(1-x)^9\, dx = \int_0^1 (1-x)x^9\, dx$

Adding: $2I = \int_0^1 x^9(1-x) + x(1-x)^9\, dx$

Easier to just expand directly:

$I = \int_0^1 x(1-x)^9\, dx = B(2,10) = \frac{1! \cdot 9!}{11!} = \frac{1}{110}$

Using Beta function: $B(m,n) = \frac{(m-1)!(n-1)!}{(m+n-1)!}$

Q3. Evaluate $\int \frac{\cos x}{(1 + \sin x)(2 + \sin x)}\, dx$

Let $t = \sin x$ , $dt = \cos x\, dx$ .

\int \frac{dt}{(1+t)(2+t)} = \int \left(\frac{1}{1+t} - \frac{1}{2+t}\right) dt = \ln|1+\sin x| - \ln|2+\sin x| + C

Q4. Evaluate $\int_0^{\pi} x \sin x\, dx$ (CBSE PYQ type)

Integration by parts: $u = x$ , $dv = \sin x\, dx$

$du = dx$ , $v = -\cos x$

[{-x\cos x}]_0^{\pi} + \int_0^{\pi} \cos x\, dx = \pi + [\sin x]_0^{\pi} = \pi + 0 = \pi

Q5. Evaluate $\int \frac{dx}{1 + \tan x}$

\int \frac{\cos x}{\cos x + \sin x}\, dx

Now apply King on definite version, or use the identity trick:

Write $\cos x = \frac{1}{2}[(\cos x + \sin x) + (\cos x - \sin x)]$

\frac{1}{2}\int 1\, dx + \frac{1}{2}\int \frac{\cos x - \sin x}{\cos x + \sin x}\, dx = \frac{x}{2} + \frac{1}{2}\ln|\sin x + \cos x| + C

Q6. If $\int_0^a f(x)\, dx = \int_0^a f(a-x)\, dx$ , evaluate $\int_0^{\pi/4} \ln(1 + \tan x)\, dx$

Let $I = \int_0^{\pi/4} \ln(1+\tan x)\, dx$

By King: $I = \int_0^{\pi/4} \ln\!\left(1 + \tan\!\left(\frac{\pi}{4} - x\right)\right) dx = \int_0^{\pi/4} \ln\!\left(\frac{2}{1+\tan x}\right) dx$

$I = \int_0^{\pi/4} \ln 2\, dx - I$

$2I = \frac{\pi}{4}\ln 2$

$I = \frac{\pi}{8}\ln 2$

Q7. Evaluate $\int e^x \left(\frac{x-1}{x^2}\right) dx$

Rewrite: $\frac{x-1}{x^2} = \frac{1}{x} - \frac{1}{x^2}$

This is exactly the form $\int e^x[f(x) + f'(x)]\, dx$ with $f(x) = \frac{1}{x}$ , $f'(x) = -\frac{1}{x^2}$ .

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

Q8. Evaluate $\int_0^{2\pi} |\sin x|\, dx$

$\sin x \geq 0$ on $[0, \pi]$ and $\sin x \leq 0$ on $[\pi, 2\pi]$ .

\int_0^{\pi} \sin x\, dx + \int_{\pi}^{2\pi} (-\sin x)\, dx = [-\cos x]_0^{\pi} + [\cos x]_{\pi}^{2\pi}

= (1+1) + (1+1) = 4

FAQs

What is the difference between definite and indefinite integrals?

An indefinite integral gives a function (family of antiderivatives) with a $+C$ . A definite integral evaluates between two limits and gives a specific number. The Fundamental Theorem of Calculus connects them: the definite integral is computed using the antiderivative evaluated at the limits.

Why do we add $+C$ in indefinite integration?

Because differentiation destroys constants — $\frac{d}{dx}(x^2 + 5) = 2x$ and $\frac{d}{dx}(x^2 + 100) = 2x$ . So $\int 2x\, dx$ could be $x^2 + 5$ or $x^2 + 100$ or any constant. We write $x^2 + C$ to capture the entire family.

Which integration method should I try first?

Check in this order: (1) Is it a standard formula directly? (2) Can substitution simplify it — do I see a function and its derivative? (3) Is it a product that fits IBP/ILATE? (4) Is it a rational function needing partial fractions? (5) Does a trig identity simplify first?

How do I handle area problems when the curve goes below the x-axis?

Find where $f(x) = 0$ in the interval. Split the integral at those zeros. Take the absolute value (or negate) the portions where $f(x) < 0$ . Total area $= \int |f(x)|\, dx$ .

What is Walli’s Formula and do I need it for JEE?

For JEE Advanced, yes. $\int_0^{\pi/2} \sin^n x\, dx = \int_0^{\pi/2} \cos^n x\, dx$ . For even $n$ : $\frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2}$ . For odd $n$ : $\frac{(n-1)!!}{n!!}$ . It saves 3-4 minutes on problems with high powers of $\sin$ or $\cos$ .

Can the limits of a definite integral be swapped?

Yes — swapping limits changes the sign: $\int_a^b f(x)\, dx = -\int_b^a f(x)\, dx$ . Also, $\int_a^a f(x)\, dx = 0$ always.

How many marks does integration carry in CBSE Class 12?

Roughly 20-22 marks across the paper — including the integrals chapter directly (around 10 marks) plus applications of integrals (area, another 6 marks) and differential equations which use integration (another 5 marks). It’s the single highest-weightage topic in Class 12 Maths.