Integral Calculus

Indefinite and definite integrals, area under curves, and reduction formulas — the highest-weightage topic in JEE Mathematics with questions appearing every year.

Indefinite Integrals Definite Integrals Area Under Curves Reduction Formulas Applications

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Indefinite Integrals

Mastering integration techniques is essential. JEE tests substitution, partial fractions, integration by parts, and recognition of standard forms.

◆ Basic Formulae & Substitution

Essential Formulae

\(\int x^n\,dx = \frac{x^{n+1}}{n+1} + C\) (\(n \neq -1\))
\(\int \frac{dx}{x} = \ln|x| + C\), \(\int e^x\,dx = e^x + C\), \(\int a^x\,dx = \frac{a^x}{\ln a} + C\)
\(\int \sin x\,dx = -\cos x + C\), \(\int \cos x\,dx = \sin x + C\)
\(\int \sec^2 x\,dx = \tan x + C\), \(\int \csc^2 x\,dx = -\cot x + C\)
\(\int \frac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\frac{x}{a} + C\), \(\int \frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a} + C\)

★ Example

Evaluate \(\int \frac{dx}{x^2 + 4x + 13}\).

Complete the square: \(x^2 + 4x + 13 = (x+2)^2 + 9\).
Let \(u = x+2\): \(\int \frac{du}{u^2 + 9} = \frac{1}{3}\tan^{-1}\frac{u}{3} + C = \frac{1}{3}\tan^{-1}\frac{x+2}{3} + C\).

◆ Partial Fractions

Decomposition Rules

Distinct linear factors: \(\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}\)
Repeated linear: \(\frac{P(x)}{(x-a)^2} = \frac{A}{x-a} + \frac{B}{(x-a)^2}\)
Irreducible quadratic: \(\frac{P(x)}{(x-a)(x^2+bx+c)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+bx+c}\)

★ Example

Evaluate \(\int \frac{2x+3}{(x+1)(x+2)}\,dx\).

\(\frac{2x+3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}\). Put \(x=-1\): \(1 = A\). Put \(x=-2\): \(-1 = -B \Rightarrow B = 1\).
\(\int \left(\frac{1}{x+1} + \frac{1}{x+2}\right)dx = \ln|x+1| + \ln|x+2| + C = \ln|(x+1)(x+2)| + C\).

◆ Integration by Parts & Special Integrals

Integration by Parts (ILATE Rule) \[\int u\,dv = uv - \int v\,du\] Choose \(u\) using ILATE priority: Inverse trig > Logarithmic > Algebraic > Trigonometric > Exponential.

JEE Special Integral \[\int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C\] Shortcut: Whenever you see \(e^x\) multiplied by a function plus its derivative, the answer is immediate.

★ Example

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

Here \(f(x) = \frac{1}{x}\) and \(f'(x) = -\frac{1}{x^2}\). So \(f(x) + f'(x) = \frac{1}{x} - \frac{1}{x^2}\).
By the special formula: \(\int e^x\left(\frac{1}{x} - \frac{1}{x^2}\right)dx = \frac{e^x}{x} + C\).

Definite Integrals

Properties of definite integrals drastically simplify JEE problems. Leibniz rule and Walli's formula are must-know tools.

◆ Properties of Definite Integrals

Key Properties (JEE Essentials)

\(\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx\) (King's Rule)
\(\displaystyle\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a-x)\,dx\)
If \(f(-x) = f(x)\) (even): \(\displaystyle\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx\)
If \(f(-x) = -f(x)\) (odd): \(\displaystyle\int_{-a}^a f(x)\,dx = 0\)
\(\displaystyle\int_0^{nT} f(x)\,dx = n\int_0^T f(x)\,dx\) if \(f\) is periodic with period \(T\)

★ Example

Evaluate \(\displaystyle\int_0^{\pi/2}\frac{\sin^3 x}{\sin^3 x + \cos^3 x}\,dx\).

Let \(I = \int_0^{\pi/2}\frac{\sin^3 x}{\sin^3 x + \cos^3 x}\,dx\).
Using King's rule (\(x \to \frac{\pi}{2} - x\)): \(I = \int_0^{\pi/2}\frac{\cos^3 x}{\cos^3 x + \sin^3 x}\,dx\).
Adding: \(2I = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}\).
\(I = \frac{\pi}{4}\).

◆ Leibniz Rule & Walli's Formula

Leibniz Integral Rule If \(F(x) = \int_{g(x)}^{h(x)} f(t)\,dt\), then: \[F'(x) = f(h(x)) \cdot h'(x) - f(g(x)) \cdot g'(x)\]

Walli's Formula \[\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\cos^n x\,dx = \begin{cases}\frac{(n-1)!!}{n!!} \cdot \frac{\pi}{2} & n \text{ even} \\ \frac{(n-1)!!}{n!!} & n \text{ odd}\end{cases}\] where \(n!! = n(n-2)(n-4)\cdots\).

★ Example

Evaluate \(\int_0^{\pi/2}\sin^4 x\,dx\).

\(n = 4\) (even). \(\frac{3!!}{4!!} \cdot \frac{\pi}{2} = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3}{8} \cdot \frac{\pi}{2} = \frac{3\pi}{16}\).

Area Under Curves

Finding areas bounded by curves is a staple of JEE. Sketch the region first, identify intersection points, then set up the integral carefully.

◆ Area Between Curves

Area Formulae

Between \(f(x)\) and \(g(x)\): \(A = \int_a^b |f(x) - g(x)|\,dx\)
Between \(x = f(y)\) and \(x = g(y)\): \(A = \int_c^d |f(y) - g(y)|\,dy\)
Key: Always identify which curve is "above" in the interval. Split the integral if curves cross.

★ Example

Find the area enclosed between \(y = x^2\) and \(y = x\).

Intersection: \(x^2 = x \Rightarrow x(x-1) = 0 \Rightarrow x = 0, 1\).
On \([0,1]\): \(x \geq x^2\). So:
\(A = \int_0^1(x - x^2)\,dx = \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1 = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}\).

◆ Standard Areas (Quick Reference)

Frequently Used Results

Area of ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\): \(\pi ab\)
Area between parabola \(y^2=4ax\) and its latus rectum: \(\frac{8a^2}{3}\)
Area enclosed by \(|x|+|y| \leq a\): \(2a^2\)

Reduction Formulas

Reduction formulas express an integral in terms of a simpler integral of the same type, reducing the power or complexity step by step.

◆ Key Reduction Formulas

Standard Reduction Formulas

\(I_n = \int \sin^n x\,dx = -\frac{\sin^{n-1}x\cos x}{n} + \frac{n-1}{n}I_{n-2}\)
\(I_n = \int \cos^n x\,dx = \frac{\cos^{n-1}x\sin x}{n} + \frac{n-1}{n}I_{n-2}\)
\(I_n = \int \tan^n x\,dx = \frac{\tan^{n-1}x}{n-1} - I_{n-2}\)
\(I_n = \int x^n e^x\,dx = x^n e^x - nI_{n-1}\)

★ Example

Evaluate \(\int_0^{\pi/2}\sin^6 x\,dx\) using the reduction formula.

\(I_6 = \frac{5}{6}I_4 = \frac{5}{6} \cdot \frac{3}{4}I_2 = \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2}I_0\).
\(I_0 = \int_0^{\pi/2}1\,dx = \frac{\pi}{2}\).
\(I_6 = \frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2} = \frac{15}{48} \cdot \frac{\pi}{2} = \frac{5\pi}{32}\).
(Matches Walli's formula for even \(n\).)

Applications — Volume of Revolution

Volume of solids of revolution using disk/washer method appears in JEE Advanced.

◆ Disk & Shell Methods

Volume Formulae

Disk method (rotation about x-axis): \(V = \pi\int_a^b [f(x)]^2\,dx\)
Washer method: \(V = \pi\int_a^b \left([R(x)]^2 - [r(x)]^2\right)dx\)
Shell method (rotation about y-axis): \(V = 2\pi\int_a^b x\,f(x)\,dx\)

★ Example

Find the volume generated by revolving \(y = \sqrt{x}\) from \(x = 0\) to \(x = 4\) about the x-axis.

\(V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = \pi \cdot 8 = 8\pi\).

★ Key Takeaways

The \(e^x[f(x)+f'(x)]\) formula saves enormous time — spot it quickly in JEE problems.
King's Rule \(\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx\) is the single most useful property of definite integrals.
For area problems, always sketch the curves and find all intersection points before integrating.
Walli's formula gives instant answers for \(\int_0^{\pi/2}\sin^n x\,dx\) without any computation.
Leibniz rule is essential for differentiating integrals with variable limits — appears frequently in JEE Advanced.

📝 Practice Problems

Problem 1

Evaluate \(\int \frac{x^2+1}{x^4+1}\,dx\).

Show Solution ▼

Divide numerator and denominator by \(x^2\): \(\frac{1+1/x^2}{x^2+1/x^2} = \frac{1+1/x^2}{(x-1/x)^2+2}\). Let \(t = x - 1/x\), \(dt = (1+1/x^2)dx\). Then \(\int \frac{dt}{t^2+2} = \frac{1}{\sqrt{2}}\tan^{-1}\frac{t}{\sqrt{2}} + C = \frac{1}{\sqrt{2}}\tan^{-1}\frac{x^2-1}{x\sqrt{2}} + C\).

Problem 2

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

Show Solution ▼

Using King's rule: \(I = \int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2 x}\,dx\). Adding: \(2I = \pi\int_0^\pi \frac{\sin x}{1+\cos^2 x}\,dx\). Let \(t = \cos x\): \(2I = \pi\int_{-1}^{1}\frac{dt}{1+t^2} = \pi[\tan^{-1}t]_{-1}^{1} = \pi\left(\frac{\pi}{4}+\frac{\pi}{4}\right) = \frac{\pi^2}{2}\). So \(I = \frac{\pi^2}{4}\).

Problem 3

Find the area of the region bounded by \(y = |x-1|\) and \(y = 3 - |x|\).

Show Solution ▼

The curves intersect where \(|x-1| = 3-|x|\). Solving in each region: for \(x \geq 1\): \(x-1 = 3-x \Rightarrow x=2\). For \(0 \leq x < 1\): \(1-x = 3-x \Rightarrow 1=3\) (no solution). For \(x < 0\): \(1-x = 3+x \Rightarrow x=-1\). Points: \((-1, 2)\) and \((2, 1)\). Area \(= \int_{-1}^{2}[(3-|x|)-|x-1|]\,dx\). Split into \([-1,0]\), \([0,1]\), \([1,2]\) and evaluate: \(A = \frac{3}{2} + 2 + \frac{1}{2} = 4\).

Problem 4

If \(f(x) = \int_0^x \frac{dt}{\sqrt{1+t^3}}\), find \(f'(2)\).

Show Solution ▼

By the Fundamental Theorem of Calculus (Leibniz rule with constant lower limit): \(f'(x) = \frac{1}{\sqrt{1+x^3}}\). So \(f'(2) = \frac{1}{\sqrt{1+8}} = \frac{1}{3}\).

Problem 5

Evaluate \(\int_0^1 x(1-x)^{10}\,dx\).

Show Solution ▼

Let \(u = 1-x\), \(du = -dx\). When \(x=0\), \(u=1\); when \(x=1\), \(u=0\). \(\int_1^0 (1-u)u^{10}(-du) = \int_0^1(u^{10}-u^{11})du = \frac{1}{11} - \frac{1}{12} = \frac{1}{132}\).

🎯 Interactive Quiz

1. \(\int e^x(\sin x + \cos x)\,dx\) equals:

A \(e^x\sin x + C\)

B \(e^x\cos x + C\)

C \(\frac{e^x}{2}(\sin x - \cos x) + C\)

D \(\frac{e^x}{2}(\sin x + \cos x) + C\)

2. \(\displaystyle\int_{-\pi/2}^{\pi/2}\frac{x\sin x}{1+\cos x}\,dx\) is evaluated using which property?

A Even function property

B King's Rule \(\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx\)

C Periodicity property

D Leibniz rule

3. The area enclosed by \(y^2 = 4x\) and \(x^2 = 4y\) is:

A \(\frac{4}{3}\)

B \(\frac{8}{3}\)

C \(\frac{16}{3}\)

D \(\frac{32}{3}\)

4. \(\displaystyle\int_0^1 \frac{\ln(1+x)}{1+x^2}\,dx\) equals:

A \(\frac{\pi}{8}\ln 2\)

B \(\frac{\pi}{4}\ln 2\)

C \(\ln 2\)

D \(\frac{\pi}{16}\ln 2\)

5. \(\displaystyle\int_0^{\pi/2}\sin^5 x\,dx\) equals:

A \(\frac{3\pi}{16}\)

B \(\frac{8}{15}\)

C \(\frac{16}{15}\)

D \(\frac{4}{15}\)