Special Definite Integrals: Must-Memorize Results for JEE

Introduction

Certain definite integrals appear so frequently in competitive exams that memorizing their results can save crucial time and mental effort. This guide presents four fundamental definite integrals with their standard results, along with proofs, related formulas, and problem-solving strategies tailored for JEE Main and Advanced.

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1. The Gaussian Integral

Standard Result

$\int_0^\infty e^{-x^2} \, dx = \frac{\sqrt{\pi}}{2}$ Full version: $\int_{-\infty}^\infty e^{-x^2} \, dx = \sqrt{\pi}$

Why it's Important

• No elementary antiderivative exists.
• Foundational in probability (normal distribution), physics (heat equation), and statistics.
• Frequently appears indirectly in JEE problems after a substitution.

Proof (Polar Coordinates)

Let $I = \int_0^\infty e^{-x^2} dx$.
Then: $I^2 = \left( \int_0^\infty e^{-x^2} dx \right) \left( \int_0^\infty e^{-y^2} dy \right) = \int_0^\infty \int_0^\infty e^{-(x^2 + y^2)} dx \, dy.$ Switch to polar coordinates: $x = r \cos\theta,\; y = r \sin\theta$, $dx \, dy = r \, dr \, d\theta$, with $r \in [0,\infty)$, $\theta \in [0, \pi/2]$: $I^2 = \int_0^{\pi/2} \int_0^\infty e^{-r^2} \, r \, dr \, d\theta.$ Let $u = r^2$, so $du = 2r \, dr$. Then: $\int_0^\infty r e^{-r^2} dr = \frac12 \int_0^\infty e^{-u} du = \frac12.$ Thus: $I^2 = \int_0^{\pi/2} \frac12 \, d\theta = \frac12 \cdot \frac{\pi}{2} = \frac{\pi}{4}.$ Hence $I = \frac{\sqrt{\pi}}{2}$.

Generalizations to Memorize

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2. Integral of $\ln(\sin x)$ and $\ln(\cos x)$

Standard Result

$\int_0^{\pi/2} \ln(\sin x) \, dx = \int_0^{\pi/2} \ln(\cos x) \, dx = -\frac{\pi}{2} \ln 2.$

Proof

Let $I = \int_0^{\pi/2} \ln(\sin x) dx$.
By the substitution $x \to \frac{\pi}{2} - x$, we get $I = \int_0^{\pi/2} \ln(\cos x) dx$.

Now: $2I = \int_0^{\pi/2} [\ln(\sin x) + \ln(\cos x)] \, dx = \int_0^{\pi/2} \ln\!\left( \frac{\sin 2x}{2} \right) dx.$ Thus: $2I = \int_0^{\pi/2} \ln(\sin 2x) \, dx - \int_0^{\pi/2} \ln 2 \, dx.$ Let $t = 2x$ in the first integral: $\int_0^{\pi/2} \ln(\sin 2x) dx = \frac12 \int_0^\pi \ln(\sin t) dt.$ Using symmetry $\sin(\pi - t) = \sin t$: $\int_0^\pi \ln(\sin t) dt = 2 \int_0^{\pi/2} \ln(\sin t) dt = 2I.$ Therefore: $2I = \frac12 (2I) - \frac{\pi}{2} \ln 2 \quad\Rightarrow\quad I = -\frac{\pi}{2} \ln 2.$

Related Results (Memorize)

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3. The $x \cdot f(\sin x)$ Property

Standard Result

For any function $f$ such that the integrals exist: $\int_0^\pi x \, f(\sin x) \, dx = \frac{\pi}{2} \int_0^\pi f(\sin x) \, dx.$

Proof

Let $I = \int_0^\pi x f(\sin x) dx$.
Replace $x$ by $\pi - x$: $I = \int_0^\pi (\pi - x) f(\sin(\pi - x)) dx = \int_0^\pi (\pi - x) f(\sin x) dx.$ Hence: $I = \pi \int_0^\pi f(\sin x) dx - I \quad\Rightarrow\quad 2I = \pi \int_0^\pi f(\sin x) dx.$ So: $I = \frac{\pi}{2} \int_0^\pi f(\sin x) dx.$

Important Applications

Example 1: Evaluate $\int_0^\pi \frac{x \sin x}{1 + \cos^2 x} dx$.
Here $f(\sin x) = \frac{\sin x}{1 + \cos^2 x}$. Thus: $I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1 + \cos^2 x} dx.$ Substitute $t = \cos x$, $dt = -\sin x \, dx$: $I = \frac{\pi}{2} \int_{1}^{-1} \frac{-dt}{1+t^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{dt}{1+t^2} = \frac{\pi}{2} \big[ \tan^{-1}t \big]_{-1}^{1} = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4}.$

Example 2: Evaluate $\int_0^\pi \frac{x}{1 + \sin x} dx$.
Here $f(\sin x) = \frac{1}{1+\sin x}$. Then: $I = \frac{\pi}{2} \int_0^\pi \frac{dx}{1+\sin x}.$ Multiply numerator and denominator by $1 - \sin x$: $I = \frac{\pi}{2} \int_0^\pi \frac{1 - \sin x}{\cos^2 x} dx = \frac{\pi}{2} \int_0^\pi (\sec^2 x - \sec x \tan x) dx = \frac{\pi}{2} \big[ \tan x - \sec x \big]_0^\pi = \pi.$

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4. The Dilogarithm Integral

Standard Result

$\int_0^1 \frac{\ln(1+x)}{x} \, dx = \frac{\pi^2}{12}.$

Proof (Series Expansion)

Recall the Taylor series: $\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n-1} x^n}{n}.$ Thus: $\frac{\ln(1+x)}{x} = \sum_{n=1}^\infty \frac{(-1)^{n-1} x^{n-1}}{n}.$ Integrate term by term from 0 to 1: $\int_0^1 \frac{\ln(1+x)}{x} dx = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \int_0^1 x^{n-1} dx = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}.$ This is the alternating sum $1 - \frac{1}{2^2} + \frac{1}{3^2} - \frac{1}{4^2} + \cdots$.
We know the Basel sum $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$.

Separate odd and even terms: $\sum_{n=1}^\infty \frac{1}{n^2} = \sum_{k=1}^\infty \frac{1}{(2k-1)^2} + \frac14 \sum_{k=1}^\infty \frac{1}{k^2}.$ Hence: $\frac{\pi^2}{6} = S_{\text{odd}} + \frac{\pi^2}{24} \quad\Rightarrow\quad S_{\text{odd}} = \frac{\pi^2}{8}.$ Now: $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2} = S_{\text{odd}} - \frac14 \sum_{k=1}^\infty \frac{1}{k^2} = \frac{\pi^2}{8} - \frac{\pi^2}{24} = \frac{\pi^2}{12}.$

Related Logarithmic Integrals

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5. Application to Previous Year JEE Problems

Problem 1 (JEE Advanced 2012)

Evaluate $\displaystyle \int_{-\pi/2}^{\pi/2} \left( x^2 + \ln\!\left(\frac{\pi+x}{\pi-x}\right) \right) \cos x \, dx$.

Solution:
Split the integral: $I = \underbrace{\int_{-\pi/2}^{\pi/2} x^2 \cos x \, dx}_{I_1} \;+\; \underbrace{\int_{-\pi/2}^{\pi/2} \ln\!\left(\frac{\pi+x}{\pi-x}\right) \cos x \, dx}_{I_2}.$

• $I_1$: The integrand $x^2 \cos x$ is even, so $I_1 = 2 \int_0^{\pi/2} x^2 \cos x \, dx.$ Use integration by parts twice (or known result) to get $I_1 = \frac{\pi^2}{2} - 4$.
• $I_2$: The function $\ln\!\left(\frac{\pi+x}{\pi-x}\right)$ is odd, and $\cos x$ is even, so their product is odd. Hence $I_2 = 0$.

Thus $I = \frac{\pi^2}{2} - 4$.

Problem 2 (JEE Main Pattern)

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

Solution:
Use the $x f(\sin x)$ property: $I = \frac{\pi}{2} \int_0^\pi \frac{\sin x}{1 + \cos^2 x} dx.$ Substitute $t = \cos x$: $I = \frac{\pi}{2} \int_{1}^{-1} \frac{-dt}{1+t^2} = \frac{\pi}{2} \int_{-1}^{1} \frac{dt}{1+t^2} = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4}.$

Problem 3 (JEE Advanced Pattern)

Prove $\displaystyle \int_0^{\pi/4} \ln(1+\tan x) \, dx = \frac{\pi}{8} \ln 2$.

Solution:
Let $I = \int_0^{\pi/4} \ln(1+\tan x) dx$.
Substitute $x \to \frac{\pi}{4} - x$: $I = \int_0^{\pi/4} \ln\!\left(1 + \tan\!\left(\frac{\pi}{4} - x\right)\right) dx = \int_0^{\pi/4} \ln\!\left(1 + \frac{1 - \tan x}{1 + \tan x}\right) dx = \int_0^{\pi/4} \ln\!\left(\frac{2}{1+\tan x}\right) dx.$ Thus: $I = \int_0^{\pi/4} \ln 2 \, dx - I \quad\Rightarrow\quad 2I = \frac{\pi}{4} \ln 2 \quad\Rightarrow\quad I = \frac{\pi}{8} \ln 2.$

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6. Quick Reference Table (Memorize!)

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7. Tips for JEE Aspirants

1. Commit these results to memory – they are often required as intermediate steps in time‑sensitive exams.
2. Recognize disguised forms – many integrals can be transformed into one of these four by a simple substitution.
3. Combine with other properties – use symmetry, periodicity, and substitution to reduce an unfamiliar integral to a standard form.
4. Practice derivations – understanding the proofs helps you adapt the results to variations that may appear in the exam.
5. Use the quick‑reference table during revision to reinforce recall.

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Master these four special definite integrals thoroughly. They are not just formulas but powerful tools that, when recognized, can turn a lengthy problem into a one‑line solution.