Properties of Definite Integrals: The Complete JEE Method

Ritesh Raj · 23 Aug 2026

23 min read18 sectionsConcept

A definite integral is a number. That single sentence decides most of what follows: a number can be estimated, compared, bounded, split, reflected and paired, and none of those moves need the antiderivative that a beginner reaches for first.

Most JEE questions on this chapter are built on that gap. They hand you an integrand whose antiderivative does not exist in closed form, then reward you for noticing a symmetry. $\int_{\pi/24}^{5\pi/24}\frac{dx}{1+\sqrt[3]{\tan 2x}}$ has no elementary antiderivative. It still takes about forty seconds.

Definite integral as signed area under a curve, with the area above the axis counted positive and below negative

Fundamental concept of the definite integral

Everything in this chapter follows from one distinction. Read this table across, not down.

	Indefinite integral $\int f(x)\,dx$	Definite integral $\int_a^b f(x)\,dx$
1. Definition	The family of all antiderivatives, $\phi(x)+C$ where $\phi'=f$	The single number $\phi(b)-\phi(a)$
2. What it produces	A function of $x$	A real number, no $x$ left in it
3. Role of the variable	$x$ survives into the answer	$x$ is a dummy: $\int_a^b f(x)\,dx=\int_a^b f(t)\,dt$
4. Constant of integration	Essential, the answer is wrong without it	Cancels in the subtraction, never write it
5. Geometrical meaning	None on its own	Signed area between the curve and the $x$ -axis from $a$ to $b$
6. Effect of swapping limits	Not applicable	Sign flips: $\int_a^b=-\int_b^a$

The word signed in row 5 is where marks are lost. Area is never negative; a definite integral often is. Where the curve dips below the axis the strip contributes a negative amount, so $\int_0^{2\pi}\sin x\,dx=0$ even though the curve encloses a total area of $4$ .

Signed area: the region above the axis counted positive, the region below counted negative, and the two cancelling for sin x over one full period

The first trap of the chapter

"Area under the curve" and "the definite integral" are not synonyms. If a question asks for area, you integrate $|f(x)|$ , which means splitting at every root of $f$ . If it asks for the integral, you do not split. Reading the wrong one costs the whole question.

The six properties, in one table

	Property	Statement	What it is really for
I	Dummy variable	$\int_a^b f(x)\,dx=\int_a^b f(t)\,dt$	Lets you rename before pairing two integrals
II	Interchanging limits	$\int_a^b f(x)\,dx=-\int_b^a f(x)\,dx$	Fixes orientation after a substitution
III	Splitting	$\int_a^b f=\int_a^c f+\int_c^b f$	Handles modulus, greatest integer, piecewise
IV	King	$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$	The workhorse of the chapter
V	Doubling on $[0,2a]$	$\int_0^{2a}f=\int_0^a f(x)\,dx+\int_0^a f(2a-x)\,dx$	Collapses to $0$ or $2\int_0^a f$
VI	Odd and even	$\int_{-a}^{a}f=0$ if $f$ odd, $=2\int_0^a f$ if $f$ even	Kills half the integrand on sight

Property III does not require $c$ to lie between $a$ and $b$ . That is worth remembering, because it lets you write $\int_{-4}^{2}=\int_{-4}^{-1}+\int_{-1}^{2}$ and also $\int_1^2=\int_1^5+\int_5^2$ when the second form is the one you can compute.

Property IV, the king property

\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx

Why it is true

Put $t=a+b-x$ , so $dt=-dx$ . When $x=a$ , $t=b$ ; when $x=b$ , $t=a$ .

\int_a^b f(a+b-x)\,dx=\int_b^a f(t)\,(-dt)=\int_a^b f(t)\,dt

and by Property I that last integral is $\int_a^b f(x)\,dx$ .

What it means geometrically

$x\mapsto a+b-x$ is the reflection of the interval $[a,b]$ in its own midpoint $\frac{a+b}{2}$ . So the property says: flip the graph left-to-right about the midpoint of the interval and the area does not change. That is obvious once you see it, and it is the whole content of the property.

Reflection of a curve about the midpoint of the interval of integration, showing the two shaded regions have equal area

The pattern that makes it pay

King is worth using when $f(x)+f(a+b-x)$ is simpler than either piece. Two shapes account for almost every JEE question:

\text{If }f(x)+f(a+b-x)=k\text{, then }\int_a^b f(x)\,dx=\frac{k(b-a)}{2}

\text{If }f(x)=\frac{g(x)}{g(x)+g(a+b-x)}\text{, then }\int_a^b f(x)\,dx=\frac{b-a}{2}

The second is the first in disguise, with $k=1$ .

How to spot it in three seconds

Add the two limits. If $a+b$ is a round number like $\frac{\pi}{2}$ , $\pi$ , $1$ , $5$ or $2\pi$ , and the integrand contains $\tan$ , $\cot$ , $\sin$ , $\cos$ or a ratio of the form $\frac{g}{g+h}$ , try king before anything else.

The one statement that is false

This is the classic two-statement question, and it has been asked in almost exactly this form more than once.

Example 1. (IIT JEE 2013) Statement I: $\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\tan x}}=\frac{\pi}{6}$ . Statement II: $\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$ . Which is correct?

▸ SHOW SOLUTION

Step 1: Statement II is Property IV, so it is true

Nothing to check.

Step 2: Apply it to Statement I

Here $a+b=\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$ , and $\tan\left(\frac{\pi}{2}-x\right)=\cot x$ .

I=\int_{\pi/6}^{\pi/3}\frac{dx}{1+\sqrt{\cot x}}=\int_{\pi/6}^{\pi/3}\frac{\sqrt{\tan x}\,dx}{1+\sqrt{\tan x}}

Step 3: Add the two forms

2I=\int_{\pi/6}^{\pi/3}\frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}}\,dx=\int_{\pi/6}^{\pi/3}dx=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}

Step 4: Read the last line carefully

$2I=\frac{\pi}{6}$ , so $I=\frac{\pi}{12}$ , not $\frac{\pi}{6}$ .

Statement I is false, Statement II is true.

The trap is not the mathematics. It is stopping at $2I$ and copying that number down. Every king-property question ends with a division by two, and the wrong option is always waiting there.

Three ways king goes wrong

Forgetting the halving. $2I=k(b-a)$ is not the answer.
Using the wrong reflection. The map is $x\mapsto a+b-x$ , built from the actual limits. On $\int_{\pi/6}^{\pi/3}$ it is $x\mapsto\frac{\pi}{2}-x$ , not $x\mapsto\pi-x$ .
Applying it across a discontinuity. King is a substitution, and a substitution needs the integrand to be integrable on the whole interval. $\int_{-1}^{1}\frac{dx}{x^2}$ is not $2\int_0^1\frac{dx}{x^2}$ ; both sides are meaningless, because the integrand blows up at $0$ .

Odd, even and the two symmetries

Property VI is the special case of Property IV in which the midpoint is the origin. It is worth separating the two symmetries a question can hand you, because students routinely apply the first when only the second is available.

A. Symmetry about zero

On an interval $[-a,a]$ :

\int_{-a}^{a}f(x)\,dx=\begin{cases}0 & \text{if }f(-x)=-f(x)\\[4pt]2\displaystyle\int_0^a f(x)\,dx & \text{if }f(-x)=f(x)\end{cases}

The value of this is not that it saves arithmetic. It is that it lets you delete terms before you integrate anything. In a sum $f=f_{\text{odd}}+f_{\text{even}}$ over a symmetric interval, the odd part contributes nothing at all, so you never compute it.

An odd function on a symmetric interval, with the two shaded regions equal in size and opposite in sign

Example 2. (JEE Main 2026, 21 Jan Shift 1) Find $\int_{-\pi/6}^{\pi/6}\frac{\pi+4x^{11}}{1-\sin\left(|x|+\frac{\pi}{6}\right)}\,dx$ .

▸ SHOW SOLUTION

Step 1: Split the numerator

The denominator $1-\sin\left(|x|+\frac{\pi}{6}\right)$ contains $|x|$ , so it is an even function of $x$ .

Step 2: Classify each piece

$\dfrac{\pi}{1-\sin\left(|x|+\frac\pi6\right)}$ is even. $\dfrac{4x^{11}}{1-\sin\left(|x|+\frac\pi6\right)}$ is odd over even, so it is odd.

Step 3: Delete the odd part

I=\int_{-\pi/6}^{\pi/6}\frac{\pi\,dx}{1-\sin\left(|x|+\frac{\pi}{6}\right)}=2\pi\int_0^{\pi/6}\frac{dx}{1-\sin\left(x+\frac{\pi}{6}\right)}

Step 4: Substitute $u=x+\frac{\pi}{6}$

I=2\pi\int_{\pi/6}^{\pi/3}\frac{du}{1-\sin u}

Step 5: Rationalise

\frac{1}{1-\sin u}=\frac{1+\sin u}{\cos^2u}=\sec^2u+\sec u\tan u

Step 6: Integrate

I=2\pi\Big[\tan u+\sec u\Big]_{\pi/6}^{\pi/3}=2\pi\left[\left(\sqrt3+2\right)-\left(\frac{1}{\sqrt3}+\frac{2}{\sqrt3}\right)\right]

Step 7: Simplify

$\frac{1}{\sqrt3}+\frac{2}{\sqrt3}=\frac{3}{\sqrt3}=\sqrt3$ , so the bracket is $\sqrt3+2-\sqrt3=2$ .

I=4\pi

The same idea, with a harder disguise, was IIT JEE 2004: $\int_{-\pi/3}^{\pi/3}\frac{\pi+4x^3}{2-\cos\left(|x|+\frac{\pi}{3}\right)}\,dx$ . Twenty-two years apart, same question.

B. Symmetry about the midpoint

When the interval is not centred at the origin, odd and even are unavailable and king is what you have. On $[0,2a]$ Property V splits it cleanly:

\int_0^{2a}f(x)\,dx=\begin{cases}0 & \text{if }f(2a-x)=-f(x)\\[4pt]2\displaystyle\int_0^{a}f(x)\,dx & \text{if }f(2a-x)=f(x)\end{cases}

Read $f(2a-x)=f(x)$ as "the graph is symmetric about the vertical line $x=a$ ", and $f(2a-x)=-f(x)$ as "the graph has a half-turn centre at $(a,0)$ ".

Why these are the same theorem

Shift by $a$ . Setting $g(u)=f(u+a)$ turns $[0,2a]$ into $[-a,a]$ , and $f(2a-x)=f(x)$ becomes $g(-u)=g(u)$ . Odd and even about the origin, and the two symmetries about $x=a$ , are one statement seen from two positions.

The boundary case: when the integrand breaks

Everything above assumed the integrand is continuous. The interesting questions are the ones where it is not, and the fix is always Property III: split the interval at every point where the formula changes, then integrate each piece with its own formula.

Three kinds of break come up:

Break	Where to split	Typical integrand
Modulus	At every root of the inside expression	$\lvert\sin x\rvert$ , $\lvert x^2+2x-3\rvert$
Greatest integer	At every integer value of the inside expression	$[x]$ , $[2\sin x]$ , $[x\sin x]$
Piecewise definition	At every changeover point given in the definition	$f(x)=\begin{cases}\cdots\end{cases}$

A step function and a modulus function on the same axes, with the split points marked on the x-axis

The split points are not always the obvious ones

For $\int_{-2}^{2}[x\sin x]\,dx$ the split is not at the integers $-1,0,1$ . It is where $x\sin x$ crosses an integer, and on $[0,2]$ that happens once, at the root of $x\sin x=1$ . Find where the inside changes value, not where $x$ does.

Type 1, king property

Example 3. (JEE Main 2026, 23 Jan Shift 1) Evaluate $\int_{\pi/24}^{5\pi/24}\frac{dx}{1+\sqrt[3]{\tan 2x}}$ .

▸ SHOW SOLUTION

Step 1: Check the limit sum

$a+b=\frac{\pi}{24}+\frac{5\pi}{24}=\frac{6\pi}{24}=\frac{\pi}{4}$ , so the reflection is $x\mapsto\frac{\pi}{4}-x$ .

Step 2: See what it does to the integrand

$2\left(\frac{\pi}{4}-x\right)=\frac{\pi}{2}-2x$ , and $\tan\left(\frac{\pi}{2}-2x\right)=\cot 2x$ .

I=\int_{\pi/24}^{5\pi/24}\frac{dx}{1+\sqrt[3]{\cot 2x}}=\int_{\pi/24}^{5\pi/24}\frac{\sqrt[3]{\tan 2x}\,dx}{1+\sqrt[3]{\tan 2x}}

Step 3: Add

2I=\int_{\pi/24}^{5\pi/24}dx=\frac{5\pi}{24}-\frac{\pi}{24}=\frac{\pi}{6}

Step 4: Halve

I=\frac{\pi}{12}

Example 4. (JEE Main 2026, 6 Apr Shift 1) Find $\int_{-\pi/4}^{\pi/4}\frac{32\cos^4x}{1+e^{\sin x}}\,dx$ .

▸ SHOW SOLUTION

Step 1: Recognise the $\frac{1}{1+e^{g}}$ shape

The reflection here is $x\mapsto-x$ , and $\sin(-x)=-\sin x$ , so the exponential inverts while $\cos^4x$ is untouched.

I=\int_{-\pi/4}^{\pi/4}\frac{32\cos^4x}{1+e^{-\sin x}}\,dx=\int_{-\pi/4}^{\pi/4}\frac{32\,e^{\sin x}\cos^4x}{1+e^{\sin x}}\,dx

Step 2: Add the two forms

2I=\int_{-\pi/4}^{\pi/4}32\cos^4x\cdot\frac{1+e^{\sin x}}{1+e^{\sin x}}\,dx=\int_{-\pi/4}^{\pi/4}32\cos^4x\,dx

Step 3: Use evenness

2I=64\int_0^{\pi/4}\cos^4x\,dx

Step 4: Reduce the power

$\cos^4x=\left(\frac{1+\cos2x}{2}\right)^2=\frac{1}{4}\left(1+2\cos2x+\cos^22x\right)=\frac{3}{8}+\frac{\cos2x}{2}+\frac{\cos4x}{8}$

Step 5: Integrate

\int_0^{\pi/4}\cos^4x\,dx=\left[\frac{3x}{8}+\frac{\sin2x}{4}+\frac{\sin4x}{32}\right]_0^{\pi/4}=\frac{3\pi}{32}+\frac14+0

Step 6: Finish

$2I=64\left(\frac{3\pi}{32}+\frac14\right)=6\pi+16$ , so $I=3\pi+8$ .

The $1+a^{g(x)}$ family

Whenever the denominator is $1+a^{g(x)}$ with $g$ odd about the midpoint, king turns the integrand into its own complement and the exponential vanishes completely. It works for $1+e^{\sin x}$ , $1+\pi^x$ , $1+2^{\tan x}$ , $1+a^x$ alike, and the base never appears in the answer. Seeing that the base is irrelevant is the fastest way to confirm you are on the right track.

Example 5. (JEE Main 2026, 4 Apr Shift 2) Evaluate $\int_0^1\cot^{-1}\left(1+x+x^2\right)dx$ .

▸ SHOW SOLUTION

Step 1: Turn the $\cot^{-1}$ into a difference

$1+x+x^2=1+x(1+x)$ , and the identity $\cot^{-1}(1+uv)=\tan^{-1}u-\tan^{-1}v$ holds when $uv>-1$ . With $u=1+x$ , $v=x$ :

\cot^{-1}\left(1+x+x^2\right)=\tan^{-1}(1+x)-\tan^{-1}x

Step 2: Split

I=\int_0^1\tan^{-1}(1+x)\,dx-\int_0^1\tan^{-1}x\,dx

Step 3: Shift the first integral

Put $u=1+x$ in the first: it becomes $\int_1^2\tan^{-1}u\,du$ . So

I=\int_1^2\tan^{-1}x\,dx-\int_0^1\tan^{-1}x\,dx

Step 4: Use the antiderivative

$\int\tan^{-1}x\,dx=x\tan^{-1}x-\frac12\log_e\left(1+x^2\right)$ . Call it $F(x)$ .

I=\big(F(2)-F(1)\big)-\big(F(1)-F(0)\big)=F(2)-2F(1)+F(0)

Step 5: Evaluate

$F(0)=0$ , $F(1)=\frac{\pi}{4}-\frac12\log_e2$ , $F(2)=2\tan^{-1}2-\frac12\log_e5$ .

I=2\tan^{-1}2-\frac12\log_e5-\frac{\pi}{2}+\log_e2

Step 6: Combine the logs

$-\frac12\log_e5+\log_e2=-\frac12\log_e5+\frac12\log_e4=\frac12\log_e\frac45=-\frac12\log_e\frac54$ .

I=2\tan^{-1}2-\frac12\log_e\left(\frac54\right)-\frac{\pi}{2}

Example 6. (IIT JEE 2011) Find $\int_{\sqrt{\log_e2}}^{\sqrt{\log_e3}}\frac{x\sin x^2}{\sin x^2+\sin\left(\log_e6-x^2\right)}\,dx$ .

▸ SHOW SOLUTION

Step 1: Kill the outer $x$ first

Put $x^2=t$ , so $2x\,dx=dt$ . The limits become $\log_e2$ and $\log_e3$ .

I=\frac12\int_{\log_e2}^{\log_e3}\frac{\sin t\,dt}{\sin t+\sin\left(\log_e6-t\right)}

Step 2: Check the limit sum

$\log_e2+\log_e3=\log_e6$ , which is exactly the constant inside the second sine. That is the signal.

Step 3: Apply king

I=\frac12\int_{\log_e2}^{\log_e3}\frac{\sin\left(\log_e6-t\right)dt}{\sin\left(\log_e6-t\right)+\sin t}

Step 4: Add

2I=\frac12\int_{\log_e2}^{\log_e3}dt=\frac12\left(\log_e3-\log_e2\right)=\frac12\log_e\frac32

Step 5: Halve

I=\frac14\log_e\left(\frac32\right)

Example 7. (IIT JEE 2002) Evaluate $\int_0^{\pi/2}\frac{x\sin x\cos x}{\cos^4x+\sin^4x}\,dx$ .

▸ SHOW SOLUTION

Step 1: King with $a+b=\frac{\pi}{2}$

Under $x\mapsto\frac{\pi}{2}-x$ the fraction $\frac{\sin x\cos x}{\cos^4x+\sin^4x}$ is unchanged, only the leading $x$ becomes $\frac{\pi}{2}-x$ .

I=\int_0^{\pi/2}\frac{\left(\frac{\pi}{2}-x\right)\sin x\cos x}{\cos^4x+\sin^4x}\,dx

Step 2: Add and solve for $I$

2I=\frac{\pi}{2}\int_0^{\pi/2}\frac{\sin x\cos x}{\cos^4x+\sin^4x}\,dx

Step 3: Divide top and bottom by $\cos^4x$

2I=\frac{\pi}{2}\int_0^{\pi/2}\frac{\tan x\sec^2x}{1+\tan^4x}\,dx

Step 4: Substitute $u=\tan^2x$ , $du=2\tan x\sec^2x\,dx$

2I=\frac{\pi}{4}\int_0^{\infty}\frac{du}{1+u^2}=\frac{\pi}{4}\cdot\frac{\pi}{2}=\frac{\pi^2}{8}

Step 5: Halve

I=\frac{\pi^2}{16}

Example 8. (JEE Main 2026, 8 Apr Shift 2) If $\int_{\pi/6}^{\pi/4}\left(\cot\left(x-\frac{\pi}{3}\right)\cot\left(x+\frac{\pi}{3}\right)+1\right)dx=\alpha\log_e\left(\sqrt3-1\right)$ , find $9\alpha^2$ .

▸ SHOW SOLUTION

Step 1: Write the product as a single quotient

\cot A\cot B+1=\frac{\cos A\cos B+\sin A\sin B}{\sin A\sin B}=\frac{\cos(A-B)}{\sin A\sin B}

With $A=x-\frac{\pi}{3}$ and $B=x+\frac{\pi}{3}$ , $A-B=-\frac{2\pi}{3}$ , so $\cos(A-B)=-\frac12$ .

Step 2: Turn the denominator into a difference of cosines

$2\sin A\sin B=\cos(A-B)-\cos(A+B)=-\frac12-\cos2x$ .

\cot A\cot B+1=\frac{-\frac12}{\frac12\left(-\frac12-\cos2x\right)}=\frac{-1}{-\frac12-\cos2x}=\frac{2}{1+2\cos2x}

Step 3: Integrate with $t=\tan x$

Using $\cos 2x=\frac{1-t^2}{1+t^2}$ , the integrand becomes $\frac{2\left(1+t^2\right)}{3-t^2}$ and $dx=\frac{dt}{1+t^2}$ :

I=\int_{1/\sqrt3}^{1}\frac{2\,dt}{3-t^2}=\frac{2}{2\sqrt3}\left[\log_e\left|\frac{\sqrt3+t}{\sqrt3-t}\right|\right]_{1/\sqrt3}^{1}

Step 4: Put in the limits

At $t=1$ : $\frac{\sqrt3+1}{\sqrt3-1}$ . At $t=\frac{1}{\sqrt3}$ : $\frac{\sqrt3+\frac1{\sqrt3}}{\sqrt3-\frac1{\sqrt3}}=\frac{4/\sqrt3}{2/\sqrt3}=2$ .

I=\frac{1}{\sqrt3}\left[\log_e\frac{\sqrt3+1}{\sqrt3-1}-\log_e2\right]

Step 5: Simplify the bracket

$\frac{\sqrt3+1}{\sqrt3-1}=\frac{\left(\sqrt3+1\right)^2}{2}=2+\sqrt3$ , so the bracket is $\log_e\left(\frac{2+\sqrt3}{2}\right)$ .

Now rationalise the other way: $\frac{1}{\left(\sqrt3-1\right)^2}=\frac{1}{4-2\sqrt3}=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}$ . So the bracket is $\log_e\left(\sqrt3-1\right)^{-2}=-2\log_e\left(\sqrt3-1\right)$ .

Step 6: Read off $\alpha$

I=-\frac{2}{\sqrt3}\log_e\left(\sqrt3-1\right)\ \Rightarrow\ \alpha=-\frac{2}{\sqrt3},\qquad 9\alpha^2=9\cdot\frac43=12

Type 2, odd and even

Example 9. (JEE Main 2026, 6 Apr Shift 2) Find $\int_{-1}^{1}\frac{x^3+|x|+1}{x^2+2|x|+1}\,dx$ .

▸ SHOW SOLUTION

Step 1: Note the denominator is even

$x^2+2|x|+1=\left(|x|+1\right)^2$ depends only on $|x|$ , so it is even.

Step 2: Split the numerator into odd and even parts

$\frac{x^3}{\left(|x|+1\right)^2}$ is odd. $\frac{|x|+1}{\left(|x|+1\right)^2}=\frac{1}{|x|+1}$ is even.

Step 3: Delete the odd part and double the even part

I=2\int_0^1\frac{dx}{x+1}=2\Big[\log_e(x+1)\Big]_0^1=2\log_e2

Example 10. (IIT JEE 2012) Evaluate $\int_{-\pi/2}^{\pi/2}\left[x^2+\log_e\left(\frac{\pi-x}{\pi+x}\right)\right]\cos x\,dx$ .

▸ SHOW SOLUTION

Step 1: Classify the two terms

$x^2\cos x$ is even. For the second, $\log_e\frac{\pi+x}{\pi-x}=-\log_e\frac{\pi-x}{\pi+x}$ , so $\log_e\frac{\pi-x}{\pi+x}$ is odd and, multiplied by the even $\cos x$ , stays odd.

Step 2: Delete the odd term

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

Step 3: Integrate by parts twice

\int x^2\cos x\,dx=x^2\sin x+2x\cos x-2\sin x

Step 4: Substitute the limits

At $x=\frac{\pi}{2}$ this is $\frac{\pi^2}{4}+0-2$ ; at $x=0$ it is $0$ .

I=2\left(\frac{\pi^2}{4}-2\right)=\frac{\pi^2}{2}-4

Example 11. (IIT JEE 2009) If $I_n=\int_{-\pi}^{\pi}\frac{\sin nx}{\left(1+\pi^x\right)\sin x}\,dx$ for $n=0,1,2,\dots$ , which of the following hold? (a) $I_n=I_{n+2}$ (b) $\sum_{m=1}^{10}I_{2m+1}=10\pi$ (c) $\sum_{m=1}^{10}I_{2m}=0$ (d) $I_n=I_{n+1}$

▸ SHOW SOLUTION

Step 1: Kill the $1+\pi^x$ with king

The interval is symmetric, so the reflection is $x\mapsto-x$ , and $\frac{\sin(-nx)}{\sin(-x)}=\frac{\sin nx}{\sin x}$ is unchanged.

I_n=\int_{-\pi}^{\pi}\frac{\pi^x\sin nx}{\left(1+\pi^x\right)\sin x}\,dx

Step 2: Add the two forms

2I_n=\int_{-\pi}^{\pi}\frac{\sin nx}{\sin x}\,dx=2\int_0^{\pi}\frac{\sin nx}{\sin x}\,dx

because $\frac{\sin nx}{\sin x}$ is even. So $I_n=\int_0^{\pi}\frac{\sin nx}{\sin x}\,dx$ .

Step 3: Build a recurrence

I_{n+2}-I_n=\int_0^{\pi}\frac{\sin(n+2)x-\sin nx}{\sin x}\,dx=\int_0^{\pi}\frac{2\cos(n+1)x\sin x}{\sin x}\,dx=2\int_0^{\pi}\cos(n+1)x\,dx=0

So $I_{n+2}=I_n$ , which is (a).

Step 4: Find the two seed values

$I_0=0$ and $I_1=\int_0^{\pi}dx=\pi$ .

Step 5: Read off the sums

Odd index gives $\pi$ every time, so $\sum_{m=1}^{10}I_{2m+1}=10\pi$ , which is (b). Even index gives $0$ every time, so $\sum_{m=1}^{10}I_{2m}=0$ , which is (c). And $I_1\ne I_2$ , so (d) fails.

Correct: (a), (b), (c).

Type 3, greatest integer and fractional part

The rule never changes: find where the inside crosses an integer, split there, and replace $[\,\cdot\,]$ by the constant it equals on each piece.

The greatest integer function drawn as a staircase, with the area under it read as a sum of rectangles

Example 12. (IIT JEE 2001) Evaluate $\int_{\pi/2}^{3\pi/2}[2\sin x]\,dx$ , where $[\,\cdot\,]$ is the greatest integer function.

▸ SHOW SOLUTION

Step 1: Track $2\sin x$ across the interval

On $\left[\frac{\pi}{2},\frac{3\pi}{2}\right]$ , $2\sin x$ decreases from $2$ down to $-2$ .

Step 2: Find the crossing points

$2\sin x=1$ at $x=\frac{5\pi}{6}$ ; $2\sin x=0$ at $x=\pi$ ; $2\sin x=-1$ at $x=\frac{7\pi}{6}$ .

Step 3: Read off the constant on each piece

[2\sin x]=\begin{cases}2,& x=\frac{\pi}{2}\\[2pt] 1,& \frac{\pi}{2}<x\le\frac{5\pi}{6}\\[2pt] 0,& \frac{5\pi}{6}<x<\pi\\[2pt] -1,& \pi\le x<\frac{7\pi}{6}\\[2pt] -2,& \frac{7\pi}{6}\le x\le\frac{3\pi}{2}\end{cases}

A single point contributes nothing to an integral, so the value $2$ can be ignored.

Step 4: Add the rectangles

I=1\cdot\frac{\pi}{3}+0\cdot\frac{\pi}{6}+(-1)\cdot\frac{\pi}{6}+(-2)\cdot\frac{\pi}{3}=\frac{\pi}{3}-\frac{\pi}{6}-\frac{2\pi}{3}=-\frac{\pi}{2}

Example 13. (JEE Main 2026, 28 Jan Shift 2) Evaluate $\int_{-\pi/2}^{\pi/2}\frac{12\left(3+[x]\right)}{3+[\sin x]+[\cos x]}\,dx$ .

▸ SHOW SOLUTION

Step 1: Tabulate the three step functions

On $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ the break points are $-1$ , $0$ and $1$ .

Interval	$[x]$	$[\sin x]$	$[\cos x]$	Integrand
$\left[-\frac{\pi}{2},-1\right)$	$-2$	$-1$	$0$	$\frac{12(1)}{2}=6$
$[-1,0)$	$-1$	$-1$	$0$	$\frac{12(2)}{2}=12$
$[0,1)$	$0$	$0$	$0$	$\frac{12(3)}{3}=12$
$\left[1,\frac{\pi}{2}\right]$	$1$	$0$	$0$	$\frac{12(4)}{3}=16$

At $x=0$ , $\cos x=1$ so $[\cos x]=1$ , but that is one point and contributes nothing.

Step 2: Multiply each value by its interval length

I=6\left(\frac{\pi}{2}-1\right)+12(1)+12(1)+16\left(\frac{\pi}{2}-1\right)

Step 3: Collect

I=\left(6+16\right)\left(\frac{\pi}{2}-1\right)+24=22\cdot\frac{\pi}{2}-22+24=11\pi+2

Example 14. (JEE Main 2026, 22 Jan Shift 1) Evaluate $\int_{-\pi/2}^{\pi/2}\frac{dx}{[x]+4}$ .

▸ SHOW SOLUTION

Step 1: List the pieces

$\frac{\pi}{2}\approx1.5708$ , so the break points inside the interval are $-1$ , $0$ and $1$ .

Interval	$[x]$	Integrand	Length
$\left[-\frac{\pi}{2},-1\right)$	$-2$	$\frac12$	$\frac{\pi}{2}-1$
$[-1,0)$	$-1$	$\frac13$	$1$
$[0,1)$	$0$	$\frac14$	$1$
$\left[1,\frac{\pi}{2}\right]$	$1$	$\frac15$	$\frac{\pi}{2}-1$

Step 2: Add

I=\left(\frac12+\frac15\right)\left(\frac{\pi}{2}-1\right)+\frac13+\frac14=\frac{7}{10}\left(\frac{\pi}{2}-1\right)+\frac{7}{12}

Step 3: Simplify

I=\frac{7\pi}{20}-\frac{7}{10}+\frac{7}{12}=\frac{7}{60}\left(3\pi-6+5\right)=\frac{7}{60}\left(3\pi-1\right)

Example 15. (JEE Main 2026, 4 Apr Shift 1) If $\int_{-2}^{2}\Big(|\sin x|+[x\sin x]\Big)dx=2\left(3-\cos2\right)+\beta$ , find $\beta\sin\left(\frac{\beta}{2}\right)$ .

▸ SHOW SOLUTION

Step 1: Do the modulus part

$|\sin x|$ is even and $\sin x\ge0$ on $[0,2]$ , so

\int_{-2}^{2}|\sin x|\,dx=2\int_0^2\sin x\,dx=2\left(1-\cos2\right)

Step 2: Deal with the step part

$x\sin x$ is even, so $[x\sin x]$ is even and $\int_{-2}^{2}[x\sin x]\,dx=2\int_0^2[x\sin x]\,dx$ .

Step 3: Locate the only crossing

On $[0,2]$ , $x\sin x$ rises from $0$ to $2\sin2\approx1.819$ , so it crosses the value $1$ exactly once. Call that point $c$ , defined by $c\sin c=1$ .

Step 4: Read off the step values

$[x\sin x]=0$ on $[0,c)$ and $=1$ on $[c,2]$ , so $\int_0^2[x\sin x]\,dx=2-c$ .

Step 5: Assemble

\int_{-2}^{2}\Big(|\sin x|+[x\sin x]\Big)dx=2-2\cos2+2(2-c)=6-2\cos2-2c=2\left(3-\cos2\right)-2c

Step 6: Identify $\beta$

Comparing, $\beta=-2c$ .

Step 7: Use the defining equation of $c$

\beta\sin\left(\frac{\beta}{2}\right)=-2c\sin(-c)=2c\sin c=2\cdot1=2

The transcendental $c$ never has to be found. That is the whole point of the question.

Example 16. (JEE Main 2026, 22 Jan Shift 2) If $\alpha=\int_0^{64}\left(x^{1/3}-\left[x^{1/3}\right]\right)dx$ , find $\frac{1}{\pi}\int_0^{\alpha\pi}\frac{\sin^2\theta}{\sin^6\theta+\cos^6\theta}\,d\theta$ .

▸ SHOW SOLUTION

Step 1: Substitute to tame the cube root

Put $x=t^3$ , so $dx=3t^2\,dt$ and $t$ runs from $0$ to $4$ .

\alpha=\int_0^4\left(t-[t]\right)3t^2\,dt=3\int_0^4 t^2\{t\}\,dt

Step 2: Split at the integers

\alpha=3\sum_{n=0}^{3}\int_n^{n+1}t^2(t-n)\,dt=3\sum_{n=0}^{3}\left[\frac{t^4}{4}-\frac{nt^3}{3}\right]_n^{n+1}

Step 3: Evaluate the general term

\frac{(n+1)^4-n^4}{4}-\frac{n\left[(n+1)^3-n^3\right]}{3}=\frac{4n^3+6n^2+4n+1}{4}-\frac{n\left(3n^2+3n+1\right)}{3}

which simplifies to $\frac{n^2}{2}+\frac{2n}{3}+\frac14$ .

Step 4: Sum for $n=0,1,2,3$

$\sum n^2=14$ , $\sum n=6$ , and there are $4$ terms, so the sum is $7+4+1=12$ .

\alpha=3\times12=36

Step 5: Recognise the second integrand as periodic

$\sin^6\theta+\cos^6\theta=1-3\sin^2\theta\cos^2\theta$ and $\sin^2\theta$ are both unchanged by $\theta\mapsto\theta+\frac{\pi}{2}$ , so the integrand has period $\frac{\pi}{2}$ .

Step 6: Count the periods

$\alpha\pi=36\pi$ contains $\frac{36\pi}{\pi/2}=72$ periods, so

\int_0^{36\pi}(\cdots)\,d\theta=72\int_0^{\pi/2}\frac{\sin^2\theta\,d\theta}{1-3\sin^2\theta\cos^2\theta}

Step 7: Evaluate one period

Write the denominator as $\sin^4\theta-\sin^2\theta\cos^2\theta+\cos^4\theta$ , divide top and bottom by $\cos^4\theta$ , then put $u=\tan\theta$ so that $du=\sec^2\theta\,d\theta$ :

\int_0^{\pi/2}\frac{\sin^2\theta\,d\theta}{1-3\sin^2\theta\cos^2\theta}=\int_0^{\infty}\frac{u^2\,du}{u^4-u^2+1}=\frac{\pi}{2}

Step 8: Finish

\frac{1}{\pi}\cdot72\cdot\frac{\pi}{2}=36

Type 4, modulus: split at the sign changes

Example 17. (JEE Main 2026, 21 Jan Shift 1) Find $6\int_0^{\pi}\left|\sin3x+\sin2x+\sin x\right|dx$ .

▸ SHOW SOLUTION

Step 1: Factorise the inside

$\sin3x+\sin x=2\sin2x\cos x$ , so

\sin3x+\sin2x+\sin x=2\sin2x\cos x+\sin2x=\sin2x\left(2\cos x+1\right)

Step 2: Factorise further

$\sin2x=2\sin x\cos x$ , so the expression is $2\sin x\cos x\left(2\cos x+1\right)$ .

Step 3: Find the sign changes on $(0,\pi)$

$\sin x>0$ throughout. $\cos x=0$ at $x=\frac{\pi}{2}$ . $2\cos x+1=0$ at $x=\frac{2\pi}{3}$ .

Step 4: Determine the sign on each piece

On $\left(0,\frac{\pi}{2}\right)$ : $\cos x>0$ , $2\cos x+1>0$ , product positive. On $\left(\frac{\pi}{2},\frac{2\pi}{3}\right)$ : $\cos x<0$ , $2\cos x+1>0$ , product negative. On $\left(\frac{2\pi}{3},\pi\right)$ : $\cos x<0$ , $2\cos x+1<0$ , product positive.

Step 5: Write down the antiderivative

F(x)=-\frac{\cos3x}{3}-\frac{\cos2x}{2}-\cos x

Step 6: Assemble with the right signs

I=6\Big[\big(F(\tfrac{\pi}{2})-F(0)\big)-\big(F(\tfrac{2\pi}{3})-F(\tfrac{\pi}{2})\big)+\big(F(\pi)-F(\tfrac{2\pi}{3})\big)\Big]

Step 7: Compute the four values

$F(0)=-\frac13-\frac12-1=-\frac{11}{6}$ , $F\!\left(\frac{\pi}{2}\right)=0+\frac12-0=\frac12$ , $F\!\left(\frac{2\pi}{3}\right)=-\frac13+\frac14+\frac12=\frac{5}{12}$ , $F(\pi)=\frac13-\frac12+1=\frac56$ .

Step 8: Substitute

I=6\left[\left(\frac12+\frac{11}{6}\right)-\left(\frac{5}{12}-\frac12\right)+\left(\frac56-\frac{5}{12}\right)\right]=6\left[\frac{7}{3}+\frac{1}{12}+\frac{5}{12}\right]=6\cdot\frac{17}{6}=17

Example 18. (IIT JEE 2014) Evaluate $\int_0^{\pi}\sqrt{1+4\sin^2\frac{x}{2}-4\sin\frac{x}{2}}\,dx$ .

▸ SHOW SOLUTION

Step 1: Recognise the perfect square

$1-4\sin\frac x2+4\sin^2\frac x2=\left(1-2\sin\frac x2\right)^2$ , so the integrand is $\left|1-2\sin\frac x2\right|$ .

Step 2: Find where the inside changes sign

$1-2\sin\frac x2=0$ gives $\sin\frac x2=\frac12$ , that is $\frac x2=\frac{\pi}{6}$ , so $x=\frac{\pi}{3}$ .

Step 3: Determine the sign

On $\left(0,\frac{\pi}{3}\right)$ , $\sin\frac x2<\frac12$ , so the inside is positive. On $\left(\frac{\pi}{3},\pi\right)$ it is negative.

Step 4: Split

I=\int_0^{\pi/3}\left(1-2\sin\frac x2\right)dx+\int_{\pi/3}^{\pi}\left(2\sin\frac x2-1\right)dx

Step 5: Use the antiderivative $x+4\cos\frac x2$

I=\left[x+4\cos\frac x2\right]_0^{\pi/3}-\left[x+4\cos\frac x2\right]_{\pi/3}^{\pi}

Step 6: Evaluate

I=\left(\frac{\pi}{3}+2\sqrt3-4\right)-\left(0+4-\frac{\pi}{3}-2\sqrt3\right)=4\sqrt3-4-\frac{\pi}{3}

Example 19. (JEE Main 2026, 8 Apr Shift 2) Let $f(x)=\frac13$ for $x\le\frac{\pi}{2}$ and $f(x)=\frac{b\left(1-\sin x\right)}{\left(\pi-2x\right)^2}$ for $x>\frac{\pi}{2}$ . If $f$ is continuous at $x=\frac{\pi}{2}$ , find $\int_0^{3b-6}\left|x^2+2x-3\right|dx$ .

▸ SHOW SOLUTION

Step 1: Take the right-hand limit

Put $x=\frac{\pi}{2}+h$ . Then $1-\sin x=1-\cos h$ and $\pi-2x=-2h$ .

\lim_{h\to0}\frac{b\left(1-\cos h\right)}{4h^2}=\frac{b}{4}\cdot\frac12=\frac{b}{8}

Step 2: Match it to $\frac13$

$\frac b8=\frac13$ gives $b=\frac83$ , so the upper limit is $3b-6=8-6=2$ .

Step 3: Factorise the modulus

$x^2+2x-3=(x+3)(x-1)$ , negative on $(-3,1)$ and positive outside.

Step 4: Split at $x=1$

I=\int_0^1\left(3-2x-x^2\right)dx+\int_1^2\left(x^2+2x-3\right)dx

Step 5: Evaluate

First piece: $\left[3x-x^2-\frac{x^3}{3}\right]_0^1=3-1-\frac13=\frac53$ . Second piece: $\left[\frac{x^3}{3}+x^2-3x\right]_1^2=\left(\frac83+4-6\right)-\left(\frac13+1-3\right)=\frac23+\frac53=\frac73$ .

Step 6: Add

I=\frac53+\frac73=4

Type 5, periodic integrands

If $f$ has period $T$ , an integral over any whole number of periods does not care where it starts.

\int_a^{a+T}f(x)\,dx=\int_0^{T}f(x)\,dx\qquad\text{and}\qquad\int_0^{nT}f(x)\,dx=n\int_0^{T}f(x)\,dx

A periodic curve with three consecutive period-blocks shaded to show they carry equal area

The three facts worth memorising

\int_a^{a+nT}f(x)\,dx=n\int_0^{T}f(x)\,dx,\qquad \int_{mT}^{nT}f(x)\,dx=(n-m)\int_0^{T}f(x)\,dx

\int_{a+nT}^{b+nT}f(x)\,dx=\int_a^b f(x)\,dx

Example 20. (JEE Main 2026, 2 Apr Shift 2) Find $\int_0^{20\pi}\left(\sin^4x+\cos^4x\right)dx$ .

▸ SHOW SOLUTION

Step 1: Simplify the integrand

\sin^4x+\cos^4x=\left(\sin^2x+\cos^2x\right)^2-2\sin^2x\cos^2x=1-\frac{\sin^22x}{2}

Step 2: Find the period

$\sin^22x=\frac{1-\cos4x}{2}$ has period $\frac{2\pi}{4}=\frac{\pi}{2}$ .

Step 3: Count the periods

$20\pi\div\frac{\pi}{2}=40$ .

Step 4: Integrate over one period

\int_0^{\pi/2}\left(1-\frac{1-\cos4x}{4}\right)dx=\int_0^{\pi/2}\left(\frac34+\frac{\cos4x}{4}\right)dx=\frac34\cdot\frac{\pi}{2}+0=\frac{3\pi}{8}

Step 5: Multiply

I=40\cdot\frac{3\pi}{8}=15\pi

Example 21. (IIT JEE 2004) Show that $\int_0^{n\pi+v}\left|\sin x\right|dx=2n+1-\cos v$ , where $n$ is a positive integer and $0\le v<\pi$ .

▸ SHOW SOLUTION

Step 1: Split off the whole periods

$\left|\sin x\right|$ has period $\pi$ , so

\int_0^{n\pi+v}\left|\sin x\right|dx=\int_0^{n\pi}\left|\sin x\right|dx+\int_{n\pi}^{n\pi+v}\left|\sin x\right|dx

Step 2: Evaluate the whole-period part

\int_0^{\pi}\left|\sin x\right|dx=\int_0^{\pi}\sin x\,dx=2\ \Rightarrow\ \int_0^{n\pi}\left|\sin x\right|dx=2n

Step 3: Shift the tail back to the origin

By periodicity $\int_{n\pi}^{n\pi+v}\left|\sin x\right|dx=\int_0^{v}\left|\sin x\right|dx$ , and $\sin x\ge0$ on $[0,v]\subseteq[0,\pi)$ .

Step 4: Integrate the tail

\int_0^{v}\sin x\,dx=1-\cos v

Step 5: Add

\int_0^{n\pi+v}\left|\sin x\right|dx=2n+1-\cos v

Example 22. (JEE Main 2026, 28 Jan Shift 1) Find $\displaystyle\sum_{r=1}^{20}\left\lfloor\sqrt{\pi\left(\int_0^r x\left|\sin\pi x\right|dx\right)}\right\rfloor$ .

▸ SHOW SOLUTION

Step 1: Integrate over one unit block

On $[n,n+1]$ put $x=n+u$ . Then $\left|\sin\pi x\right|=\sin\pi u$ for $u\in[0,1]$ , so

\int_n^{n+1}x\left|\sin\pi x\right|dx=\int_0^1(n+u)\sin\pi u\,du=n\cdot\frac{2}{\pi}+\frac{1}{\pi}=\frac{2n+1}{\pi}

Step 2: Sum the blocks

\int_0^r x\left|\sin\pi x\right|dx=\sum_{n=0}^{r-1}\frac{2n+1}{\pi}=\frac{r^2}{\pi}

because the odd numbers up to $2r-1$ add to $r^2$ .

Step 3: Undo the outer operations

$\pi\cdot\frac{r^2}{\pi}=r^2$ , so $\sqrt{\ \cdot\ }=r$ and $\lfloor r\rfloor=r$ .

Step 4: Add

\sum_{r=1}^{20}r=\frac{20\cdot21}{2}=210

Type 6, the integral as an unknown constant

When a definite integral has constant limits, it is a number, so it can be given a name and solved for. This turns a functional equation into ordinary algebra.

Example 23. (JEE Main 2026, 24 Jan Shift 2) If $f(x)=e^x+\int_0^1\left(y+xe^x\right)f(y)\,dy$ , find $e+f(0)$ .

▸ SHOW SOLUTION

Step 1: Name the two constants

A=\int_0^1 y\,f(y)\,dy,\qquad B=\int_0^1 f(y)\,dy

Both are numbers, so $f(x)=e^x+A+Bxe^x$ .

Step 2: Feed the form back into $B$

B=\int_0^1\left(e^y+A+Bye^y\right)dy=(e-1)+A+B\Big[ye^y-e^y\Big]_0^1=(e-1)+A+B

Step 3: Read off $A$

The $B$ cancels, leaving $0=(e-1)+A$ , so $A=1-e$ .

Step 4: Evaluate at zero

f(0)=e^0+A+0=1+(1-e)=2-e

Step 5: Finish

e+f(0)=e+2-e=2

Note that $B$ was never needed. Whenever one of the named constants cancels, stop chasing it.

Example 24. (JEE Main 2026, 24 Jan Shift 1) A differentiable $f$ satisfies $\int_0^{36}f\!\left(\frac{tx}{36}\right)dt=4\alpha f(x)$ . If $y=f(x)$ is a standard parabola through $(2,1)$ and $(-4,\beta)$ , find $\beta^{\alpha}$ .

▸ SHOW SOLUTION

Step 1: Identify the parabola

A standard parabola of the form $y=\frac{x^2}{4a}$ through $(2,1)$ gives $1=\frac{4}{4a}$ , so $a=1$ and $f(x)=\frac{x^2}{4}$ .

Step 2: Compute the left side

\int_0^{36}\frac{1}{4}\left(\frac{tx}{36}\right)^2dt=\frac{x^2}{4\cdot36^2}\int_0^{36}t^2\,dt=\frac{x^2}{4\cdot36^2}\cdot\frac{36^3}{3}=\frac{36\,x^2}{12}=3x^2

Step 3: Compare with the right side

$3x^2=4\alpha\cdot\frac{x^2}{4}=\alpha x^2$ , so $\alpha=3$ .

Step 4: Find $\beta$

\beta=f(-4)=\frac{16}{4}=4

Step 5: Finish

\beta^{\alpha}=4^3=64

Type 7, Newton–Leibnitz inside a definite integral

When the unknown function sits inside an integral whose upper limit is $x$ , differentiate instead of integrating. That is the Newton–Leibnitz rule:

\frac{d}{dx}\int_{u(x)}^{v(x)}f(t)\,dt=f\big(v(x)\big)v'(x)-f\big(u(x)\big)u'(x)

Constant limits versus variable limits

Constant limits mean the integral is a number: name it and solve algebraically, as in Type 6. A variable upper limit means the integral is a function: differentiate it, as here. Deciding which of the two you are looking at is the entire opening move.

Example 25. (JEE Main 2026, 22 Jan Shift 1) Let $f:[1,\infty)\to\mathbb{R}$ be differentiable with $6\int_1^{x}f(t)\,dt=3xf(x)+x^3-4$ for all $x\ge1$ . Find $f(2)-f(3)$ .

▸ SHOW SOLUTION

Step 1: Differentiate both sides

6f(x)=3f(x)+3xf'(x)+3x^2

Step 2: Rearrange into a linear equation

3f(x)-3xf'(x)=3x^2\ \Rightarrow\ f'(x)-\frac{f(x)}{x}=-x

Step 3: Recognise the derivative of a quotient

\left(\frac{f}{x}\right)'=\frac{f'}{x}-\frac{f}{x^2}=\frac1x\left(f'-\frac fx\right)=-1

Step 4: Integrate

\frac{f(x)}{x}=-x+C\ \Rightarrow\ f(x)=-x^2+Cx

Step 5: Fix $C$ from the original relation at $x=1$

The left side is $0$ , so $0=3f(1)+1-4$ , giving $f(1)=1$ , hence $-1+C=1$ and $C=2$ .

Step 6: Evaluate

$f(x)=-x^2+2x$ , so $f(2)=0$ and $f(3)=-3$ .

f(2)-f(3)=3

Example 26. (JEE Main 2026, 23 Jan Shift 2) Find the number of elements of $S=\left\{x\in[0,100]:\int_0^{x}t^2\sin(x-t)\,dt=x^2\right\}$ .

▸ SHOW SOLUTION

Step 1: Turn the convolution into a plain integral

Put $u=x-t$ , so $t=x-u$ and $dt=-du$ ; the limits swap.

\int_0^{x}t^2\sin(x-t)\,dt=\int_0^{x}(x-u)^2\sin u\,du

Step 2: Expand

=x^2\int_0^x\sin u\,du-2x\int_0^x u\sin u\,du+\int_0^x u^2\sin u\,du

Step 3: Use the standard antiderivatives

$\int_0^x\sin u\,du=1-\cos x$ ; $\int_0^x u\sin u\,du=\sin x-x\cos x$ ; $\int_0^x u^2\sin u\,du=2x\sin x-\left(x^2-2\right)\cos x-2$ .

Step 4: Collect

x^2-x^2\cos x-2x\sin x+2x^2\cos x+2x\sin x-x^2\cos x+2\cos x-2=x^2-2+2\cos x

Step 5: Solve the equation

$x^2-2+2\cos x=x^2$ gives $\cos x=1$ , that is $x=2k\pi$ .

Step 6: Count them in $[0,100]$

$2k\pi\le100$ needs $k\le\frac{50}{\pi}\approx15.9$ , so $k=0,1,\dots,15$ .

\left|S\right|=16

Example 27. (JEE Main 2026, 28 Jan Shift 2) Let $f(x)=1-2x+\int_0^{x}e^{x-t}f(t)\,dt$ and $g(x)=\int_0^{x}\left(f(t)+2\right)^{15}(t-4)^6(t+12)^{17}dt$ . If $p$ and $q$ are the points of local minimum and maximum of $g$ , find $\left|p+q\right|$ .

▸ SHOW SOLUTION

Step 1: Pull the $e^x$ out

f(x)=1-2x+e^{x}\int_0^{x}e^{-t}f(t)\,dt

Step 2: Differentiate

f'(x)=-2+e^{x}\int_0^{x}e^{-t}f(t)\,dt+e^x\cdot e^{-x}f(x)=-2+\big(f(x)-1+2x\big)+f(x)

Step 3: Simplify and solve

$f'=2f+2x-3$ with $f(0)=1$ . Trying $f=\alpha x+\beta$ gives $\alpha=2\beta-3$ and $0=2\alpha+2$ , so $\alpha=-1$ , $\beta=1$ .

f(x)=1-x

Step 4: Write down $g'$

g'(x)=\left(3-x\right)^{15}(x-4)^6(x+12)^{17}

Step 5: Read the sign changes off the exponents

$(x-4)^6$ has an even exponent, so it never changes sign. $(3-x)^{15}$ flips at $x=3$ , and $(x+12)^{17}$ flips at $x=-12$ .

Step 6: Tabulate

For $x<-12$ : positive times negative, so $g'<0$ . For $-12<x<3$ : positive times positive, so $g'>0$ . For $x>3$ : negative times positive, so $g'<0$ .

Step 7: Classify

$g'$ goes negative to positive at $x=-12$ , a local minimum, so $p=-12$ . It goes positive to negative at $x=3$ , a local maximum, so $q=3$ .

\left|p+q\right|=\left|-9\right|=9

Type 8, estimating without evaluating

Some questions never want the value. They want a bound, and the tool is the comparison property:

m\le f(x)\le M\text{ on }[a,b]\ \Longrightarrow\ m(b-a)\le\int_a^b f(x)\,dx\le M(b-a)

A curve trapped between its minimum and maximum on an interval, with the two bounding rectangles drawn

Two companions are worth knowing: $g\le f\le h$ on $[a,b]$ gives $\int_a^b g\le\int_a^b f\le\int_a^b h$ , and $\left|\int_a^b f\right|\le\int_a^b\left|f\right|$ .

Example 28. (JEE Advanced 2012) Let $f'(x)=\frac{192x^3}{2+\sin^4(\pi x)}$ for all $x\in\mathbb{R}$ , with $f\!\left(\frac12\right)=0$ . If $m\le\int_{1/2}^{1}f(x)\,dx\le M$ , find the possible $m$ and $M$ .

▸ SHOW SOLUTION

Step 1: Bound the derivative

$\sin^4(\pi x)\in[0,1]$ , so $2+\sin^4(\pi x)\in[2,3]$ and, for $x>0$ ,

\frac{192x^3}{3}\le f'(x)\le\frac{192x^3}{2},\qquad\text{that is}\qquad 64x^3\le f'(x)\le96x^3

Step 2: Integrate the bounds from $\frac12$ to $x$

\int_{1/2}^{x}64t^3\,dt\le f(x)-f\!\left(\tfrac12\right)\le\int_{1/2}^{x}96t^3\,dt

Step 3: Use $f\!\left(\frac12\right)=0$

16x^4-1\le f(x)\le24x^4-\frac32

Step 4: Integrate again from $\frac12$ to $1$

\int_{1/2}^{1}\left(16x^4-1\right)dx\le\int_{1/2}^{1}f(x)\,dx\le\int_{1/2}^{1}\left(24x^4-\frac32\right)dx

Step 5: Evaluate both sides

Left: $\frac{16}{5}\left(1-\frac1{32}\right)-\frac12=\frac{16}{5}\cdot\frac{31}{32}-\frac12=\frac{31}{10}-\frac12=\frac{26}{10}=2.6$ . Right: $\frac{24}{5}\cdot\frac{31}{32}-\frac34=\frac{93}{20}-\frac34=\frac{78}{20}=3.9$ .

Step 6: Match against the options

$2.6\le\int_{1/2}^{1}f\le3.9$ , so any valid pair must satisfy $m\le2.6$ and $M\ge3.9$ . Of the offered pairs only $m=1,\ M=12$ works.

Example 29. (IIT JEE 2000) Let $g(x)=\int_0^{x}f(t)\,dt$ where $\frac12\le f(t)\le1$ for $t\in[0,1]$ and $0\le f(t)\le\frac12$ for $t\in[1,2]$ . Bound $g(2)$ .

▸ SHOW SOLUTION

Step 1: Split at the changeover

g(2)=\int_0^1 f(t)\,dt+\int_1^2 f(t)\,dt

Step 2: Bound each piece with its own constants

\frac12\le\int_0^1 f\le1,\qquad 0\le\int_1^2 f\le\frac12

Step 3: Add the inequalities

\frac12\le g(2)\le\frac32

Decision chart

Decision chart for choosing the right definite-integral property from the shape of the integrand

What you see	First move
$a+b$ is round, integrand has $\tan$ , $\cot$ or $\frac{g}{g+h}$	King, then add and halve
Denominator $1+a^{g(x)}$ with $g$ odd about the midpoint	King, the base cancels
Limits $-a$ to $a$	Split into odd and even parts, delete the odd one
Limits $0$ to $2a$	Test $f(2a-x)$ against $f(x)$
$[\,\cdot\,]$ or $\{\,\cdot\,\}$ anywhere	Find where the inside crosses an integer, then split
$\lvert\ \cdot\ \rvert$ anywhere	Find the roots of the inside, then split
Limits are a large multiple of a period	Count periods, integrate one
Integral with constant limits inside the definition of $f$	Name it, solve algebraically
Integral with variable upper limit	Differentiate, Newton–Leibnitz
Question asks for a range, not a value	Bound the integrand, then integrate the bounds

Quick recall

Situation	Result
Swap the limits	$\int_a^b f=-\int_b^a f$
King	$\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx$
$f(x)+f(a+b-x)=k$	$\int_a^b f=\frac{k(b-a)}{2}$
$f(x)=\frac{g(x)}{g(x)+g(a+b-x)}$	$\int_a^b f=\frac{b-a}{2}$
$f$ odd on $[-a,a]$	$\int_{-a}^{a}f=0$
$f$ even on $[-a,a]$	$\int_{-a}^{a}f=2\int_0^a f$
$f(2a-x)=-f(x)$	$\int_0^{2a}f=0$
$f(2a-x)=f(x)$	$\int_0^{2a}f=2\int_0^a f$
$f$ has period $T$	$\int_a^{a+nT}f=n\int_0^{T}f$
$\int_0^{n\pi+v}\lvert\sin x\rvert\,dx$	$2n+1-\cos v$
$\int_0^r x\lvert\sin\pi x\rvert\,dx$	$\frac{r^2}{\pi}$
$m\le f\le M$ on $[a,b]$	$m(b-a)\le\int_a^b f\le M(b-a)$
Newton–Leibnitz	$\frac{d}{dx}\int_{u}^{v}f(t)\,dt=f(v)v'-f(u)u'$

The habit that makes this chapter fast

Before writing a single line, do three checks in this order. One: add the limits and see whether $a+b$ is round. Two: look for $\lvert\ \cdot\ \rvert$ , $[\,\cdot\,]$ or a piecewise definition, and if one is there, find the split points first. Three: ask whether the limits are constants or contain $x$ . Those three checks classify nearly every question in this chapter, and each takes about five seconds.

What a wrong start costs you

Reaching for the antiderivative on a king-property question is not a slower route to the same answer. On $\int_{\pi/24}^{5\pi/24}\frac{dx}{1+\sqrt[3]{\tan 2x}}$ there is no antiderivative to find, so the attempt goes nowhere and takes four minutes to establish that. In an exam where a question is worth about ninety seconds, the cost of the wrong opening move is the question plus two more.

Work through the full chapter with proofs, the complete problem bank and worked solutions in the IITianForum lecture notes PDF, then come back and try the ten shifts you have not seen yet.

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Reinforce this post with live problems on doMath.

Definite Integration Application of Integrals Differential Equations

Properties of Definite Integrals: The Complete JEE Method

Fundamental concept of the definite integral

The six properties, in one table

Property IV, the king property

Why it is true

What it means geometrically

The pattern that makes it pay

The one statement that is false

Step 1: Statement II is Property IV, so it is true

Step 2: Apply it to Statement I

Step 3: Add the two forms

Step 4: Read the last line carefully

Statement I is false, Statement II is true.

Odd, even and the two symmetries

A. Symmetry about zero

Step 1: Split the numerator

Step 2: Classify each piece

Step 3: Delete the odd part

Step 4: Substitute u=x+π6u=x+\frac{\pi}{6}u=x+6π​

Step 5: Rationalise

Step 6: Integrate

Step 7: Simplify

B. Symmetry about the midpoint

The boundary case: when the integrand breaks

Type 1, king property

Step 1: Check the limit sum

Step 2: See what it does to the integrand

Step 3: Add

Step 4: Halve

Step 1: Recognise the 11+eg\frac{1}{1+e^{g}}1+eg1​ shape

Step 2: Add the two forms

Step 3: Use evenness

Step 4: Reduce the power

Step 5: Integrate

Step 6: Finish

Step 1: Turn the cot⁡−1\cot^{-1}cot−1 into a difference

Step 2: Split

Step 3: Shift the first integral

Step 4: Use the antiderivative

Step 5: Evaluate

Step 6: Combine the logs

Step 1: Kill the outer xxx first

Step 2: Check the limit sum

Step 3: Apply king

Step 4: Add

Step 5: Halve

Step 1: King with a+b=π2a+b=\frac{\pi}{2}a+b=2π​

Step 2: Add and solve for III

Step 3: Divide top and bottom by cos⁡4x\cos^4xcos4x

Step 4: Substitute u=tan⁡2xu=\tan^2xu=tan2x, du=2tan⁡xsec⁡2x dxdu=2\tan x\sec^2x\,dxdu=2tanxsec2xdx

Step 5: Halve

Step 1: Write the product as a single quotient

Step 2: Turn the denominator into a difference of cosines

Step 3: Integrate with t=tan⁡xt=\tan xt=tanx

Step 4: Put in the limits

Step 5: Simplify the bracket

Step 6: Read off α\alphaα

Type 2, odd and even

Step 1: Note the denominator is even

Step 2: Split the numerator into odd and even parts

Step 3: Delete the odd part and double the even part

Step 1: Classify the two terms

Step 2: Delete the odd term

Step 3: Integrate by parts twice

Step 4: Substitute the limits

Step 1: Kill the 1+πx1+\pi^x1+πx with king

Step 2: Add the two forms

Step 3: Build a recurrence

Step 4: Find the two seed values

Step 5: Read off the sums

Correct: (a), (b), (c).

Type 3, greatest integer and fractional part

Step 1: Track 2sin⁡x2\sin x2sinx across the interval

Step 2: Find the crossing points

Step 3: Read off the constant on each piece

Step 4: Add the rectangles

Step 1: Tabulate the three step functions

Step 2: Multiply each value by its interval length

Step 3: Collect

Step 1: List the pieces

Step 4: Substitute $u=x+\frac{\pi}{6}$

Step 1: Recognise the $\frac{1}{1+e^{g}}$ shape

Step 1: Turn the $\cot^{-1}$ into a difference

Step 1: Kill the outer $x$ first

Step 1: King with $a+b=\frac{\pi}{2}$

Step 2: Add and solve for $I$

Step 3: Divide top and bottom by $\cos^4x$

Step 4: Substitute $u=\tan^2x$ , $du=2\tan x\sec^2x\,dx$

Step 3: Integrate with $t=\tan x$

Step 6: Read off $\alpha$

Step 1: Kill the $1+\pi^x$ with king

Step 1: Track $2\sin x$ across the interval

Step 6: Identify $\beta$

Step 7: Use the defining equation of $c$

Step 4: Sum for $n=0,1,2,3$

Step 3: Find the sign changes on $(0,\pi)$

Step 5: Use the antiderivative $x+4\cos\frac x2$

Step 2: Match it to $\frac13$

Step 4: Split at $x=1$

Step 2: Feed the form back into $B$

Step 3: Read off $A$

Step 4: Find $\beta$