Integrals

The chapter runs in two large, connected halves. The first half builds the indefinite integral: it opens with the definition of an antiderivative and the standard list of integrals read straight off the derivative table (Section 7.2), states the linearity properties that let those formulas combine, and then develops the three real techniques of integration in turn — substitution and trigonometric identities (7.3), the six standard formulas for quadratic-denominator integrals built by completing the square (7.4), partial fractions for rational functions (7.5), and integration by parts for products, including the special $e^x[f(x)+f'(x)]$ shortcut and the $\sqrt{x^2\pm a^2}$, $\sqrt{a^2-x^2}$ family (7.6). The second half turns to the definite integral, introduced first as a signed area (7.7), then unlocked by the Fundamental Theorem of Calculus in its two parts — the area function's derivative equals the integrand, and the evaluation shortcut $F(b)-F(a)$ (7.8). Section 7.9 shows how to evaluate a definite integral by substitution, most efficiently by changing the limits along with the variable. The chapter closes with eight properties of definite integrals, P0 through P7 (7.10), which use symmetry — reversing limits, splitting the interval, or replacing $x$ by $a+b-x$ — to evaluate integrals that would otherwise need no antiderivative to exist in closed form at all.

Antiderivatives and the indefinite integral

Differential calculus hands you a function and asks for its derivative. Integral calculus reverses the question: given a derivative, find the function. If $F$ is any function with $\dfrac{d}{dx}F(x)=f(x)$ on an interval $I$, then $F$ is called an antiderivative (or primitive) of $f$. Because the derivative of a constant is zero, $F+C$ is an antiderivative of $f$ for every real number $C$, and it can be proved (by the mean value theorem, sketched in the first derivation below) that these are the only antiderivatives — any two functions with the same derivative on an interval differ by a constant. The entire family is bundled into one notation, the indefinite integral:

where: $f(x)$ is the integrand, $x$ is the variable of integration, and $C$ is the constant of integration — an arbitrary real number, never to be dropped, since two different values of $C$ give genuinely different curves in the family. Reading formula (1) as a statement about derivatives, differentiating an indefinite integral undoes it, $\dfrac{d}{dx}\int f(x)\,dx = f(x)$, and integrating a derivative recovers the function up to a constant, $\int f'(x)\,dx = f(x)+C$ — the two together are the working definition of "inverse process." From the standard derivative table, a list of standard integrals follows immediately: $\int x^n\,dx = \dfrac{x^{n+1}}{n+1}+C$ for $n\neq -1$; $\int \dfrac{1}{x}\,dx = \log|x|+C$ (the case $n=-1$, handled separately because it needs the absolute value to be valid for negative $x$ too); $\int \cos x\,dx=\sin x+C$; $\int \sin x\,dx=-\cos x+C$; $\int \sec^2x\,dx=\tan x+C$; $\int \operatorname{cosec}^2x\,dx=-\cot x+C$; $\int \sec x\tan x\,dx=\sec x+C$; $\int \operatorname{cosec}x\cot x\,dx=-\operatorname{cosec}x+C$; $\int \dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+C$; $\int \dfrac{dx}{1+x^2}=\tan^{-1}x+C$; $\int e^x\,dx=e^x+C$; and $\int a^x\,dx=\dfrac{a^x}{\log a}+C$. These formulas are the alphabet every other integral in the chapter is spelled with. Integration is linear, which is what makes term-by-term integration legitimate:

where: $k$ is any real constant, and both rules generalise to any finite combination $\int[k_1f_1(x)+\cdots+k_nf_n(x)]\,dx = k_1\int f_1(x)\,dx+\cdots+k_n\int f_n(x)\,dx$. Finding an antiderivative by pattern-matching against the standard list is called integration by inspection; for example, since $\dfrac{d}{dx}(x^3+x^4)=3x^2+4x^3$, an antiderivative of $3x^2+4x^3$ is $x^3+x^4$, found purely by recognising the derivative. Inspection is fast but limited — most integrands need one of the three systematic techniques below.

Integration by substitution

When an integrand contains a function together with (a constant multiple of) its own derivative, changing the variable strips that structure away. Given $I=\int f(x)\,dx$, substitute $x=g(t)$, so that $dx=g'(t)\,dt$:

where: the substitution is chosen so the new integral in $t$ is a standard form. The commonest pattern is $t=$ (the "inner" function), $dt=$ (its derivative) $dx$: for $\int 2x\sin(x^2+1)\,dx$, put $t=x^2+1$, $dt=2x\,dx$, so the integral becomes $\int \sin t\,dt=-\cos t+C=-\cos(x^2+1)+C$. Trigonometric identities often prepare an integrand for substitution or direct standard-form matching — writing $\cos^2x=\frac{1+\cos2x}{2}$, or $\sin x\cos y=\frac12[\sin(x+y)+\sin(x-y)]$, or $\sin^3x=\frac{3\sin x-\sin3x}{4}$, converts a power or product of trigonometric functions into a sum of terms that integrate directly. Four integrals worth memorising as standard results, each derivable by substitution, are $\int \tan x\,dx=\log|\sec x|+C$, $\int \cot x\,dx=\log|\sin x|+C$, $\int \sec x\,dx=\log|\sec x+\tan x|+C$, and $\int \operatorname{cosec}x\,dx=\log|\operatorname{cosec}x-\cot x|+C$.

Integrals of some particular functions

A second family of integrals — those with a quadratic in the denominator, or under a square root — is handled by reducing to six master formulas:

where: the first two come from partial fractions, the middle pair from the substitution $x=a\tan\theta$ or $x=a\sec\theta$, and the last two from $x=a\sin\theta$ or $x=a\tan\theta$ (the first derivation below proves the first and fifth in full). Any quadratic $ax^2+bx+c$ that does not already look like $x^2\pm a^2$ or $a^2-x^2$ is forced into that shape by completing the square: $ax^2+bx+c = a\left[\left(x+\dfrac{b}{2a}\right)^2 + \left(\dfrac{c}{a}-\dfrac{b^2}{4a^2}\right)\right]$, after which the substitution $t=x+\dfrac{b}{2a}$ reduces the integral to one of the six formulas. When the numerator is linear, $px+q$, it is split as $px+q = A\cdot\dfrac{d}{dx}(ax^2+bx+c) + B = A(2ax+b)+B$, matching coefficients to find $A$ and $B$; the $A$-part integrates by the substitution $t=ax^2+bx+c$ (giving a logarithm), and the $B$-part reduces to one of formulas (4)–(5) after completing the square.

Integration by partial fractions

A rational function $P(x)/Q(x)$ is proper if the degree of $P$ is less than the degree of $Q$; an improper rational function is first reduced to a polynomial plus a proper part by long division, $\dfrac{P(x)}{Q(x)} = T(x) + \dfrac{P_1(x)}{Q(x)}$, and $T(x)$ integrates termwise as a polynomial. A proper rational function, in turn, is rewritten as a sum of simpler fractions according to the shape of its denominator: for distinct linear factors, $\dfrac{px+q}{(x-a)(x-b)} = \dfrac{A}{x-a}+\dfrac{B}{x-b}$; for a repeated linear factor, $\dfrac{px+q}{(x-a)^2} = \dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$; for three distinct linear factors, three separate terms $\dfrac{A}{x-a}+\dfrac{B}{x-b}+\dfrac{C}{x-c}$; for a repeated factor alongside a simple one, $\dfrac{px^2+qx+r}{(x-a)^2(x-b)} = \dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}+\dfrac{C}{x-b}$; and when an irreducible quadratic sits alongside a linear factor, $\dfrac{px^2+qx+r}{(x-a)(x^2+bx+c)} = \dfrac{A}{x-a}+\dfrac{Bx+C}{x^2+bx+c}$, where $x^2+bx+c$ cannot be factored further over the reals.

where: in every pattern, the unknown constants $A$, $B$, $C$ are found by clearing denominators to get a polynomial identity, then either equating coefficients of matching powers of $x$ on both sides, or substituting convenient values of $x$ (such as $x=a$ or $x=b$, which kill all but one term at once). Once decomposed, each piece is a standard form: $\int \dfrac{dx}{x-a}=\log|x-a|+C$ for a simple linear factor, and $\int \dfrac{dx}{(x-a)^2}=-\dfrac{1}{x-a}+C$ for a repeated one, while the irreducible-quadratic piece is split further (as in the previous subsection) into a logarithmic part and an inverse-tangent part.

Integration by parts

For products of functions that inspection cannot crack, integration by parts unwinds the product rule of differentiation. If $u=f(x)$ is the first function and $\dfrac{dv}{dx}=g(x)$ gives the second function, then

where: in words — the integral of a product equals (first function) times (integral of the second) minus the integral of [(derivative of the first) times (integral of the second)]. The choice of which factor is "first" is decisive: choosing badly turns a solvable integral into a harder one (as happens if $\cos x$ is taken first in $\int x\cos x\,dx$, which increases the power of $x$ instead of removing it). The standard priority, remembered by the acronym ILATE — Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential — picks the first function as whichever type appears earliest in that list; algebraic functions like $x^n$ are usually taken first over trigonometric or exponential partners, but inverse-trigonometric and logarithmic functions, which have no elementary antiderivative of their own to guess by inspection, are always taken first, paired with the constant function $1$ when nothing else is present (as in $\int \log x\,dx = x\log x - x + C$). A named shortcut saves repeated work whenever the integrand has the shape $e^x[f(x)+f'(x)]$:

where: this follows from applying (6) once, with $f(x)$ as the first function and $e^x$ as the second, and watching the two leftover integrals cancel exactly — it is proved in full in the second derivation below and reused constantly, since it converts what looks like integration by parts into a one-line pattern match. A further pair of standard results, proved by taking the constant function $1$ as the second function, gives closed forms for $\int \sqrt{x^2\pm a^2}\,dx$ and $\int \sqrt{a^2-x^2}\,dx$; these too are self-referential, and the technique of solving for the unknown integral algebraically after it reappears on the right-hand side is the same one used for $\int e^x\sin x\,dx$.

The definite integral and the Fundamental Theorem of Calculus

A definite integral $\int_a^b f(x)\,dx$, with lower limit $a$ and upper limit $b$, is a specific real number — geometrically, the area of the region bounded by $y=f(x)$, the ordinates $x=a$ and $x=b$, and the $x$-axis (formally introduced as a limit of Riemann sums, though this chapter's working tool is the theorem below rather than the sum). Define the area function $A(x)=\int_a^x f(t)\,dt$, the running area from $a$ up to a variable point $x$ inside $[a,b]$. Two theorems, together called the Fundamental Theorem of Calculus, connect this area function to antiderivatives.

where: $F$ is any antiderivative of $f$ (continuous on $[a,b]$) — not necessarily the area function $A$ itself. The practical procedure, proved in the third derivation below, is always the same: find one antiderivative $F(x)$ of $f(x)$, ignoring the constant $C$ (it cancels automatically, since $[F(x)+C]_a^b=[F(b)+C]-[F(a)+C]=F(b)-F(a)$), then evaluate $F(b)-F(a)$. The Second FTC is what makes definite integrals computable without ever summing a rectangle, and it is why the entire indefinite-integral toolkit built earlier in the chapter feeds directly into area calculations.

Evaluating definite integrals by substitution

When a definite integral is evaluated by the substitution method, the fastest route changes the limits of integration along with the variable, so there is no need to substitute back to $x$ at the end: given $\int_a^b f(x)\,dx$ with substitution $t=g(x)$, find the new limits $t=g(a)$ and $t=g(b)$, integrate the transformed integrand in $t$, and evaluate directly at the new limits. For $\int_{-1}^{1} 5x^4\sqrt{x^5+1}\,dx$, putting $t=x^5+1$ gives $dt=5x^4\,dx$, and the limits transform from $x=-1,1$ to $t=0,2$, so the whole computation stays in $t$ until the final subtraction — no resubstitution to $x^5+1$ is ever needed.

Properties of definite integrals

Eight properties, labelled P0 through P7, use symmetry to evaluate definite integrals that direct antidifferentiation cannot reach in closed form (a striking number of JEE and board problems are solved only by one of these):

where: $P_0$ says the name of the dummy variable is irrelevant; $P_1$ says reversing the limits flips the sign; $P_2$ splits the interval at any interior point $c$; $P_3$ and $P_4$ replace $x$ by $a+b-x$ (or $a-x$, when the lower limit is $0$) without changing the integral's value — the workhorse of definite-integral tricks, proved in the fourth derivation below, and used by adding the original and transformed integrals to eliminate an awkward $x$ that appears explicitly in the integrand; $P_5$ and $P_6$ extend the same idea to an interval of length $2a$; and $P_7$ is the even/odd shortcut, either halving the work over a symmetric interval or annihilating the integral outright.

Board level - completing the square to a standard inverse-trigonometric form

Evaluate $\displaystyle\int \frac{dx}{x^2-6x+13}$.

Step 1. Complete the square in the denominator: $x^2-6x+13 = x^2-6x+3^2-3^2+13 = (x-3)^2+4 = (x-3)^2+2^2$.

Step 2. Substitute $t=x-3$, so $dt=dx$, reducing the integral to the standard form $\displaystyle\int \frac{dt}{t^2+2^2}$, which matches formula (4) with $a=2$.

Step 3. Apply the standard formula $\int \dfrac{dt}{t^2+a^2}=\dfrac1a\tan^{-1}\dfrac{t}{a}+C$ directly: $\displaystyle\int \frac{dt}{t^2+2^2}=\frac12\tan^{-1}\frac{t}{2}+C$. Answer: $\displaystyle\int \frac{dx}{x^2-6x+13}=\frac{1}{2}\tan^{-1}\!\left(\frac{x-3}{2}\right)+C$.

Board level - a definite integral evaluated by changing the limits under substitution

Evaluate $\displaystyle\int_{-1}^{1} 5x^4\sqrt{x^5+1}\,dx$.

Step 1. Substitute $t=x^5+1$, so $dt=5x^4\,dx$ — exactly the combination sitting in the integrand — and change the limits immediately: when $x=-1$, $t=(-1)^5+1=0$; when $x=1$, $t=1^5+1=2$.

Step 2. The integral becomes entirely a $t$-integral with the new limits, no resubstitution to $x$ needed: $\displaystyle\int_0^2 \sqrt{t}\,dt = \int_0^2 t^{1/2}\,dt = \left[\frac{2}{3}t^{3/2}\right]_0^2$.

Step 3. Evaluate at the limits: $\dfrac{2}{3}(2)^{3/2} - \dfrac{2}{3}(0)^{3/2} = \dfrac{2}{3}\cdot 2\sqrt2 = \dfrac{4\sqrt2}{3}$. Answer: $\displaystyle\int_{-1}^{1} 5x^4\sqrt{x^5+1}\,dx = \dfrac{4\sqrt2}{3}$.

Competitive level (JEE / NEET) - the self-referential integration-by-parts loop

Find $\displaystyle\int e^x\sin x\,dx$.

Step 1. Take $e^x$ as the first function and $\sin x$ as the second, so $\int \sin x\,dx=-\cos x$. Integration by parts gives $I = -e^x\cos x + \int e^x\cos x\,dx = -e^x\cos x + I_1$, where $I_1=\int e^x\cos x\,dx$.

Step 2. Apply integration by parts to $I_1$ the same way — $e^x$ first, $\cos x$ second, so $\int \cos x\,dx = \sin x$ — giving $I_1 = e^x\sin x - \int e^x\sin x\,dx = e^x\sin x - I$.

Step 3. Substitute $I_1$ back: $I = -e^x\cos x + e^x\sin x - I$, so $2I = e^x(\sin x - \cos x)$, and $I = \dfrac{e^x}{2}(\sin x-\cos x)+C$. Answer: $\displaystyle\int e^x\sin x\,dx = \frac{e^x}{2}(\sin x-\cos x)+C$. Timing note: about ninety seconds once the loop is recognised. The trap is switching which function is "first" between the two rounds — keep $e^x$ first both times, since swapping regenerates the same integral with the opposite sign instead of the same integral you started with, and the algebra never closes. Recognise the pattern $e^x\times(\text{trig})$ instantly and commit to the loop rather than searching for a substitution; JEE frequently disguises this as $\int e^{2x}\sin 3x\,dx$ or similar, solved by the identical two-round trick.

Competitive level (JEE / NEET) - a definite integral collapsed by the a+b-x property

Evaluate $\displaystyle I=\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}}$.

Step 1. Apply property $P_3$ with $a=\pi/6$, $b=\pi/3$, so $a+b-x = \pi/2-x$: replace $x$ by $\pi/2-x$ throughout the integrand. Since $\tan(\pi/2-x)=\cot x$, this gives $I=\displaystyle\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\cot x}}$.

Step 2. Rationalise the new integrand: $\dfrac{1}{1+\sqrt{\cot x}} = \dfrac{1}{1+\frac{1}{\sqrt{\tan x}}} = \dfrac{\sqrt{\tan x}}{\sqrt{\tan x}+1}$, so $I = \displaystyle\int_{\pi/6}^{\pi/3} \frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\,dx$.

Step 3. Add this to the original expression for $I$: $2I = \displaystyle\int_{\pi/6}^{\pi/3}\left[\frac{1}{1+\sqrt{\tan x}}+\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\right]dx = \int_{\pi/6}^{\pi/3} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}}\,dx = \int_{\pi/6}^{\pi/3} 1\,dx = \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}$.

Step 4. Solve for $I$: $I=\dfrac{1}{2}\cdot\dfrac{\pi}{6}=\dfrac{\pi}{12}$. Answer: $\displaystyle\int_{\pi/6}^{\pi/3} \frac{dx}{1+\sqrt{\tan x}} = \frac{\pi}{12}$. Timing note: under two minutes once $P_3$ is recognised, and it is entirely algebra-free — no antiderivative of $1/(1+\sqrt{\tan x})$ is ever found, or needs to exist in closed form. The signal to look for is limits that sum to a recognisable angle ($\pi/6+\pi/3=\pi/2$ here) paired with an integrand that turns into a companion expression (here $\sqrt{\tan x}\leftrightarrow\sqrt{\cot x}$) under $x\to a+b-x$; whenever adding the original and transformed integrands produces something constant, the property will collapse the problem instantly.

Dropping the constant of integration $C$, or writing it twice. Every indefinite integral is a family of curves, not one curve; forgetting $C$ loses marks even when the "shape" of the answer is right, and after a definite integral is evaluated, $C$ must not reappear at all — it cancels automatically in $F(b)-F(a)$ and should never be written in a numerical final answer.

Losing the absolute value in logarithmic integrals. $\int \dfrac{dx}{x}=\log|x|+C$, not $\log x+C$ — the domain of $\log x$ excludes negative numbers, but $1/x$ is defined for all $x\neq 0$. The same care is needed in every partial-fraction and standard-form answer that produces a logarithm.

Choosing the wrong "first function" in integration by parts. Picking $\cos x$ as first and $x$ as second in $\int x\cos x\,dx$ increases the power of $x$ instead of eliminating it, sending the calculation in the wrong direction; the ILATE priority exists precisely to avoid this trap.

Forgetting to change the limits during substitution in a definite integral. If the variable is changed from $x$ to $t$, either change the limits to the corresponding $t$-values and finish entirely in $t$, or resubstitute back to $x$ before applying the original limits — mixing the two (new limits with the old variable, or vice versa) gives a wrong numerical answer even when every other step is correct.

Misusing the properties $P_3$/$P_4$ in the wrong direction, or on the wrong interval. $P_3$ and $P_4$ replace $x$ by $a+b-x$ (or $a-x$) only when the limits themselves are $a$ and $b$ (or $0$ and $a$) — applying the substitution with the wrong pair of numbers, or forgetting that the technique requires adding the transformed integral to the original one to make progress, is the single most common way these otherwise-powerful shortcuts are misapplied.