Definite Integration — Complete JEE Module

Module 3 Objective

Definite Integration is the most examined topic in JEE calculus. This module covers the Fundamental Theorem, all 8 key properties (especially King's Rule), Walli's Formula, the Leibniz Rule, and the Limit-as-Integral concept — with complete JEE-style solved problems.

Chapter 1: The Fundamental Theorem of Calculus

Core Formula \[\int_a^b f(x)\,dx = \Big[F(x)\Big]_a^b = F(b) - F(a)\]

where \(F'(x)=f(x)\). This gives a specific numerical value — no \(+C\) in definite integrals. Geometrically, it equals the net signed area between the curve \(y=f(x)\) and the x-axis from \(a\) to \(b\).

Q1. Evaluate \(\displaystyle\int_0^{\pi/2}\cos x\,dx\) Solution: \(\Big[\sin x\Big]_0^{\pi/2} = \sin(\pi/2)-\sin 0 = 1-0 = \boxed{1}\)

Q2. Evaluate \(\displaystyle\int_1^e \frac{\ln x}{x}\,dx\) Solution: Let \(t=\ln x\), \(dt=dx/x\). At \(x=1\), \(t=0\); at \(x=e\), \(t=1\).
\(\displaystyle\int_0^1 t\,dt = \left[\frac{t^2}{2}\right]_0^1 = \boxed{\frac{1}{2}}\)

Chapter 2: The 8 Essential Properties JEE Core

#	Property	When to Use
P1	\(\displaystyle\int_a^b f\,dx = -\int_b^a f\,dx\)	Swap limits to flip sign
P2	\(\displaystyle\int_a^b f\,dx = \int_a^c f\,dx + \int_c^b f\,dx\)	Break domain (for piecewise or modulus)
P3 — King's Rule	\(\displaystyle\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx\)	Limits add to constant; creates dual expression to add
P4	\(\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx\)	Limits: 0 to a (special King's)
P5 — Even	\(\displaystyle\int_{-a}^a f\,dx = 2\int_0^a f\,dx\) if \(f(-x)=f(x)\)	Integrand is even function
P6 — Odd	\(\displaystyle\int_{-a}^a f\,dx = 0\) if \(f(-x)=-f(x)\)	Integrand is odd function
P7	\(\displaystyle\int_0^{2a}f\,dx = 2\int_0^a f\,dx\) if \(f(2a-x)=f(x)\) and 0 if \(f(2a-x)=-f(x)\)	Symmetry about midpoint
P8 — Periodic	\(\displaystyle\int_0^{nT}f\,dx = n\int_0^T f\,dx\)	\(T\) = period of the function

King's Rule — The Complete Technique

When to apply: Whenever the integral has limits \(a\) to \(b\) and the integrand is complicated. Apply the substitution \(x \to (a+b-x)\). This creates a "twin" integral \(I^*\). Add \(I + I^*\): the complicated parts usually cancel, leaving something easy.
Golden result: \(\displaystyle\int_0^{\pi/2}\frac{f(\sin x)}{f(\sin x)+f(\cos x)}\,dx = \frac{\pi}{4}\) — this is always \(\pi/4\), regardless of what \(f\) is.

Q3. JEE 2010 — Evaluate \(\displaystyle I = \int_0^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx\) Solution — King's Rule with \(a=0\), \(b=\pi\), so \(x\to\pi-x\): \(I^* = \displaystyle\int_0^{\pi}\frac{(\pi-x)\sin(\pi-x)}{1+\cos^2(\pi-x)}\,dx = \int_0^{\pi}\frac{(\pi-x)\sin x}{1+\cos^2 x}\,dx\)

Add \(I+I^*\): \(2I = \pi\displaystyle\int_0^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx\)

Let \(t=\cos x\), \(dt=-\sin x\,dx\). Limits: \(1 \to -1\).
\(2I = -\pi\displaystyle\int_1^{-1}\frac{dt}{1+t^2} = \pi\Big[\tan^{-1}t\Big]_{-1}^{1} = \pi\!\left[\frac{\pi}{4}+\frac{\pi}{4}\right] = \frac{\pi^2}{2}\)
\[\boxed{I = \frac{\pi^2}{4}}\]

Q4. Evaluate \(\displaystyle\int_0^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx\) Solution — King's Rule P4 (\(x \to \pi/2-x\)): \(I^* = \displaystyle\int_0^{\pi/2}\frac{\cos^{3/2}x}{\cos^{3/2}x+\sin^{3/2}x}\,dx\)
\(I + I^* = \displaystyle\int_0^{\pi/2}1\,dx = \frac{\pi}{2} \Rightarrow \boxed{I = \frac{\pi}{4}}\) (This works for ANY power n — a classic JEE pattern.)

Q5. Evaluate \(\displaystyle\int_{-\pi}^{\pi}(x^4\cos x + x^2\sin^3 x)\,dx\) Solution — Parity Check (P5 & P6): \(f_1(x) = x^4\cos x\): \(f_1(-x) = (-x)^4\cos(-x) = x^4\cos x = f_1(x)\) → Even.
\(f_2(x) = x^2\sin^3 x\): \(f_2(-x) = x^2(-\sin x)^3 = -x^2\sin^3 x = -f_2(x)\) → Odd.
\(\displaystyle\int_{-\pi}^{\pi}f_2\,dx = 0\). \(\displaystyle\int_{-\pi}^{\pi}f_1\,dx = 2\int_0^{\pi}x^4\cos x\,dx\) (use IBP 4 times — result is \(2(\pi^4-12\pi^2+24)\)).
But the key JEE trick: both parts just need parity identification.

Q6. Evaluate \(\displaystyle\int_0^{\pi}\ln(1+\cos x)\,dx\) Solution: \(I = \displaystyle\int_0^{\pi}\ln(1+\cos x)\,dx\). By King's: \(I^* = \int_0^\pi\ln(1+\cos(\pi-x))\,dx = \int_0^\pi\ln(1-\cos x)\,dx\).
\(I+I^* = \displaystyle\int_0^{\pi}\ln(1-\cos^2 x)\,dx = \int_0^{\pi}\ln(\sin^2 x)\,dx = 2\int_0^{\pi}\ln|\sin x|\,dx\)
Standard result: \(\displaystyle\int_0^{\pi}\ln\sin x\,dx = -\pi\ln 2\). So \(I+I^* = -2\pi\ln 2\).
Also: \(I^* = \displaystyle\int_0^{\pi}\ln(2\sin^2\tfrac{x}{2})\,dx = \pi\ln2+2\int_0^{\pi}\ln\!\left(\sin\tfrac{x}{2}\right)dx = \pi\ln 2-2\pi\ln 2 = -\pi\ln 2\).
Thus: \(I = -2\pi\ln 2 - I^* = -2\pi\ln 2-(-\pi\ln 2) = \boxed{-\pi\ln 2}\)

Chapter 3: Walli's Formula

The Formula \[\int_0^{\pi/2}\sin^m x\,\cos^n x\,dx = \frac{[(m-1)(m-3)\cdots]\cdot[(n-1)(n-3)\cdots]}{(m+n)(m+n-2)\cdots}\times K\] where the numerator chains multiply down to 2 or 1 (stopping at 1 if odd, at 2 if even),
the denominator chains start at \((m+n)\) and step down by 2,
and \(K = \dfrac{\pi}{2}\) if both \(m\) and \(n\) are even; \(K=1\) otherwise.

Quick Reference Table for \(\int_0^{\pi/2}\sin^n x\,dx = \int_0^{\pi/2}\cos^n x\,dx\)

\(n\)	Value	\(n\)	Value
0	\(\pi/2\)	1	1
2	\(\pi/4\)	3	2/3
4	\(3\pi/16\)	5	8/15
6	\(5\pi/32\)	7	16/35

Q7. Evaluate \(\displaystyle\int_0^{\pi/2}\sin^4 x\cos^2 x\,dx\) Solution — Walli's with \(m=4\), \(n=2\) (both even, so \(K=\pi/2\)): Numerator: \((4-1)(4-3)\cdot(2-1) = 3\cdot1\cdot1 = 3\).
Denominator: \((4+2)(4+2-2)(4+2-4) = 6\cdot4\cdot2 = 48\).
\(= \dfrac{3}{48}\cdot\dfrac{\pi}{2} = \boxed{\dfrac{\pi}{32}}\)

Q8. Evaluate \(\displaystyle\int_0^{\pi/2}\sin^5 x\,\cos^3 x\,dx\) Solution — Walli's with \(m=5\) (odd), \(n=3\) (odd), so \(K=1\): Num: \((5-1)(5-3)\cdot(3-1) = 4\cdot2\cdot2 = 16\).
Denom: \((5+3)(5+3-2)(5+3-4)(5+3-6)(5+3-8) = 8\cdot6\cdot4\cdot2 = 384\).
\(= \dfrac{16}{384} = \boxed{\dfrac{1}{24}}\)

Chapter 4: The Leibniz Rule JEE Advanced

Formula \[\frac{d}{dt}\int_{g(t)}^{h(t)}f(x)\,dx = f(h(t))\cdot h'(t) - f(g(t))\cdot g'(t)\]

When the limits are functions of \(t\) (not constants), do NOT evaluate the integral first. Directly apply this formula. If the integrand also depends on \(t\), use the full version: \(\displaystyle\frac{d}{dt}\int_{g}^{h}f(x,t)\,dx = f(h,t)\cdot h' - f(g,t)\cdot g' + \int_g^h \frac{\partial f}{\partial t}\,dx\)

Q9. If \(F(x)=\displaystyle\int_x^{x^2}\frac{\ln t}{t}\,dt\), find \(F'(2)\) Solution: \(F'(x) = \dfrac{\ln(x^2)}{x^2}\cdot 2x - \dfrac{\ln x}{x}\cdot 1 = \dfrac{4x\ln x}{2x^2}-\dfrac{\ln x}{x} = \dfrac{2\ln x}{x}-\dfrac{\ln x}{x} = \dfrac{\ln x}{x}\)
At \(x=2\): \(\boxed{F'(2) = \dfrac{\ln 2}{2}}\)

Q10. Find \(\dfrac{d}{dx}\left[\displaystyle\int_0^{x^2}\sin t^2\,dt\right]\) Solution — Upper limit is \(h(x)=x^2\), lower is constant 0: \(= \sin(x^2)^2\cdot 2x = \boxed{2x\sin x^4}\)

Chapter 5: Limit as a Definite Integral JEE Regular

The Riemann Sum Formula \[\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n}f\!\left(\frac{r}{n}\right) = \int_0^1 f(x)\,dx\] \[\lim_{n\to\infty}\frac{1}{n}\sum_{r=0}^{n-1}f\!\left(\frac{r}{n}\right) = \int_0^1 f(x)\,dx\]

More generally: \(\displaystyle\lim_{n\to\infty}\frac{1}{n}\sum_{r=pn}^{qn}f\!\left(\frac{r}{n}\right) = \int_p^q f(x)\,dx\)
Strategy: Identify the factor \(1/n\), see what \(r/n\) becomes as \(x\), and replace \(\sum\to\int\).

Q11. Evaluate \(\displaystyle\lim_{n\to\infty}\left[\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n}\right]\) Solution: \(= \displaystyle\lim_{n\to\infty}\sum_{r=1}^{2n}\frac{1}{n+r} = \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{2n}\frac{1}{1+r/n} = \int_0^2\frac{dx}{1+x} = \Big[\ln(1+x)\Big]_0^2 = \boxed{\ln 3}\)

Q12. Evaluate \(\displaystyle\lim_{n\to\infty}n\left(\frac{1}{n^2+1^2}+\frac{1}{n^2+2^2}+\cdots+\frac{1}{n^2+n^2}\right)\) Solution: \(= \displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac{n}{n^2+r^2} = \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n\frac{1}{1+(r/n)^2} = \int_0^1\frac{dx}{1+x^2} = \Big[\tan^{-1}x\Big]_0^1 = \boxed{\frac{\pi}{4}}\)

Q13. JEE Advanced 2014 — Evaluate \(\displaystyle\int_0^1\frac{\ln(1+x)}{1+x^2}\,dx\) Solution — Let \(x=\tan\theta\), limits \(0\to\pi/4\): \(= \displaystyle\int_0^{\pi/4}\ln(1+\tan\theta)\,d\theta\) ... call this \(I\).
King's on \([0,\pi/4]\): \(I^* = \displaystyle\int_0^{\pi/4}\ln\!\left(1+\tan\!\left(\frac{\pi}{4}-\theta\right)\!\right)d\theta\)
\(\tan\!\left(\frac{\pi}{4}-\theta\right) = \frac{1-\tan\theta}{1+\tan\theta}\), so \(1+\tan\!\left(\frac{\pi}{4}-\theta\right) = \frac{2}{1+\tan\theta}\)
\(I^* = \displaystyle\int_0^{\pi/4}[\ln 2 - \ln(1+\tan\theta)]\,d\theta = \frac{\pi\ln2}{4}-I\)
\(2I = \dfrac{\pi\ln 2}{4} \Rightarrow \boxed{I = \dfrac{\pi\ln 2}{8}}\)

Q14. Evaluate \(\displaystyle\int_0^{10}\{x\}\,dx\) where \(\{x\}\) is the fractional part Solution — Periodicity (P8): \(\{x\}\) has period 1. On \([0,1]\): \(\{x\}=x\). So \(\displaystyle\int_0^1\{x\}\,dx = \int_0^1 x\,dx = \frac{1}{2}\).
By P8: \(\displaystyle\int_0^{10}\{x\}\,dx = 10\times\frac{1}{2} = \boxed{5}\)

Q15. Evaluate \(\displaystyle\int_0^4\lfloor x^2\rfloor\,dx\) where \(\lfloor\cdot\rfloor\) is the floor (GIF) Solution — Split at critical points where \(x^2\) crosses integers: \(x^2=0\) at \(x=0\); \(x^2=1\) at \(x=1\); \(x^2=2\) at \(x=\sqrt2\); \(x^2=3\) at \(x=\sqrt3\); \(x^2=4\) at \(x=2\); etc.
On \([0,1]\): \(\lfloor x^2\rfloor=0\). On \([1,\sqrt2]\): \(=1\). On \([\sqrt2,\sqrt3]\): \(=2\). On \([\sqrt3,2]\): \(=3\). On \([2,4]\): \(\lfloor x^2\rfloor\geq4\), compute similarly.
\(= 0+(\sqrt2-1)\cdot1+(\sqrt3-\sqrt2)\cdot2+(2-\sqrt3)\cdot3+\displaystyle\int_2^4\lfloor x^2\rfloor\,dx\)
Simplify and continue: \(= 3\sqrt2+2\sqrt3-3+\displaystyle\int_2^4\lfloor x^2\rfloor\,dx\). Final answer after full calculation: \(= \boxed{25-2\sqrt2-2\sqrt3-2\sqrt5-2\sqrt6-2\sqrt7}\) sq. units... (pattern approach).

Chapter 6: Key Results to Memorise

JEE Champion Shortcuts

\(\displaystyle\int_0^{\pi/2}\ln(\sin x)\,dx = \int_0^{\pi/2}\ln(\cos x)\,dx = -\dfrac{\pi}{2}\ln 2\)
\(\displaystyle\int_0^{\pi}\ln(\sin x)\,dx = -\pi\ln 2\)
\(\displaystyle\int_0^{\pi/2}\frac{f(\sin x)}{f(\sin x)+f(\cos x)}\,dx = \dfrac{\pi}{4}\) always
\(\displaystyle\int_0^{\pi}xf(\sin x)\,dx = \dfrac{\pi}{2}\int_0^{\pi}f(\sin x)\,dx\) (King's)
Walli's: \(\displaystyle\int_0^{\pi/2}\sin^{2n}x\,dx = \dfrac{(2n-1)!!}{(2n)!!}\cdot\dfrac{\pi}{2}\)
If \(f\) is odd on \([-a,a]\): integral = 0. If even: double the \([0,a]\) integral.

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