JEE Main Maths Integral Calculus 2027: Standard Integrals, Substitution & Area Under Curves

The NTA syllabus for this chapter literally lists out a set of standard integral forms, which is a strong hint about what gets tested. Learn to recognise each pattern on sight, and a large share of Integral Calculus questions become quick lookups rather than fresh derivations.

Integration is pattern recognition. Match the integral to a standard form, and the answer follows.

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Chapter at a Glance

Priority reflects past-paper trends, not an official NTA weightage.

What the NTA Syllabus Covers

• Integral as an anti-derivative, fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions
• Integration by substitution, by parts and by partial fractions, integration using trigonometric identities
• Standard integrals of the listed forms: $\int \frac{dx}{x^2 \pm a^2}$, $\int \frac{dx}{\sqrt{x^2 \pm a^2}}$, $\int \frac{dx}{a^2 - x^2}$, $\int \frac{dx}{\sqrt{a^2 - x^2}}$, and the related forms with quadratics $ax^2 + bx + c$ in the denominator or under a root, and $\int \sqrt{a^2 \pm x^2}\,dx$, $\int \sqrt{x^2 - a^2}\,dx$
• Fundamental theorem of calculus, properties of definite integrals, evaluation of definite integrals, areas of regions bounded by simple curves in standard forms

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Master These Topics

1. The Standard Forms the NTA Syllabus Names

$$\int \frac{dx}{x^2 + a^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C \qquad \int \frac{dx}{a^2 - x^2} = \frac{1}{2a}\ln\left\lvert\frac{a+x}{a-x}\right\rvert + C$$

$$\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac{x}{a} + C \qquad \int \frac{dx}{\sqrt{x^2 \pm a^2}} = \ln\left\lvert x + \sqrt{x^2 \pm a^2}\right\rvert + C$$

For denominators of the form $ax^2 + bx + c$, complete the square first to reduce it to one of these four forms.

Worked example: $\displaystyle\int \frac{dx}{x^2 + 4} = \frac{1}{2}\tan^{-1}\frac{x}{2} + C$.

For a numerator that is linear, $\int \dfrac{(px+q)\,dx}{ax^2+bx+c}$, split it into a multiple of the derivative of the denominator plus a constant, so that one part integrates to a logarithm and the other reduces to a standard form.

Worked example: $\displaystyle\int \frac{2x + 3}{x^2 + 3x + 7}\,dx = \ln\lvert x^2 + 3x + 7 \rvert + C$, since the numerator is exactly the derivative of the denominator.

2. Integration Techniques

• Substitution: replace a function and its derivative with a single variable.
• By parts: $\displaystyle\int u\,dv = uv - \int v\,du$, choosing $u$ using the priority order Inverse, Logarithmic, Algebraic, Trigonometric, Exponential (ILATE).
• Partial fractions: split a rational function with a factorable denominator into simpler fractions before integrating.

Worked example (by parts): $\displaystyle\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C$.

3. Definite Integrals and a Useful Property

The fundamental theorem of calculus states $\displaystyle\int_a^b f(x)\,dx = F(b) - F(a)$. One property saves significant work:

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

Worked example: $\displaystyle\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx$. Applying the property with $a = \pi/2$ replaces $\sin x$ with $\cos x$ and vice versa. Adding the original integral $I$ to its transformed version gives $2I = \displaystyle\int_0^{\pi/2} 1\,dx = \pi/2$, so $I = \pi/4$.

Trap: This symmetry trick only works cleanly with limits $0$ to $a$ (or $-a$ to $a$ for even/odd functions). Do not force it onto arbitrary limits.

4. Area Under a Curve

Worked example: The area under $y = x^2$ from $x = 0$ to $x = 3$ is $\displaystyle\int_0^3 x^2\,dx = \left[\frac{x^3}{3}\right]_0^3 = \frac{27}{3} = 9$.

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Common Traps to Avoid

• Forgetting to complete the square before matching a quadratic denominator to a standard form.
• Applying the ILATE rule backwards, which leads to an integral that never simplifies.
• Using the $\int_0^a f(x) = \int_0^a f(a-x)$ trick with the wrong limits.
• Leaving out the constant of integration in an indefinite integral.

60-Second Revision Sheet

• $\int \dfrac{dx}{x^2+a^2} = \tfrac{1}{a}\tan^{-1}\tfrac{x}{a}$; $\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\tfrac{x}{a}$
• By parts: $\int u\,dv = uv - \int v\,du$ (choose $u$ by ILATE)
• $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$
• Area under a curve: $\int_a^b y\,dx$

Your Study Plan

1. Day 1: standard integral forms and completing the square.
2. Day 2: substitution, by parts and partial fractions.
3. Day 3: definite integral properties and symmetry tricks.
4. Day 4: area under curves and a timed mixed set.

Practice Integral Calculus Questions Free → (opens in a new tab)

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Continue Your Maths Journey

• Previous chapter: Limit, Continuity & Differentiability
• Next chapter: Differential Equations
• All 14 JEE Main Maths chapters
• Complete JEE Main Syllabus 2027 guide

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