Integrals

1. Check this before you revise anything

The formal "definite integral as a limit of a sum" derivation is gone. Older editions built the definite integral from scratch — partition $[a,b]$ into $n$ strips of width $h=(b-a)/n$, sum $f$ at each partition point, take the limit as $h\to 0$ — as its own worked section with its own exercise.

The current book keeps only a single sentence: "the definite integral is introduced either as the limit of a sum or if it has an antiderivative F in the interval $[a,b]$, then its value is the difference between the values of F at the end points." It even says outright, while introducing the properties in section 7.10, that they let you evaluate "without calculating the limit of a sum."

None of the 261 questions across this chapter's 10 exercises and Miscellaneous Exercise ever ask you to build a definite integral from a raw partition sum. Every one is solved through the Fundamental Theorem of Calculus — find an antiderivative, evaluate it at the two limits, subtract — or through the properties of definite integrals in section 7.10.

The old stub had no exercises at all — a generic formula table (basic forms, "ILATE," a one-line description of partial fractions) with no solutions file behind it. The real book has ten numbered exercises plus a Miscellaneous Exercise:

261 questions in total — the largest chapter in the Class 12 Mathematics book, more than double the next-largest (Continuity and Differentiability, 131).

The textbook's own section map, for navigation:

2. Integration as the Inverse of Differentiation (Textbook 7.2)

Differentiation takes a function to its derivative. Integration asks the reverse question: given $f$, which function has $f$ as its derivative?

A function $F$ with $F^{\prime }(x)=f(x)$ is called an antiderivative or primitive of $f$.

Why the constant appears. If $F$ is one antiderivative, so is $F+C$ for every constant $C$, since the derivative of a constant is zero. There is no way to distinguish them from the derivative alone, so the answer is a whole family:

$\int f(x)dx=F(x)+C$

The symbol $\int$ denotes the indefinite integral, $f$ is the integrand, $C$ the constant of integration. Omitting $C$ is treated as an error in indefinite-integral answers.

Reading the derivative tables backwards gives the basic list that Exercise 7.1 tests entirely by inspection:

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C(n\ne -1),\int \frac{1}{x}dx=\log |x|+C,\int e^{x}dx=e^x+C$

$\int \cos xdx=\sin x+C,\int \sin xdx=-\cos x+C,\int {\sec }^{2}xdx=\tan x+C$

Note the two exceptions students most often miss: $n=-1$ is excluded from the power rule and handled by the logarithm, and the modulus in $\log |x|$ is required because $x$ may be negative.

Properties (7.2.1). Integration is linear:

$\int \left[f(x)\pm g(x)\right]dx=\int f(x)dx\pm \int g(x)dx,\int kf(x)dx=k\int f(x)dx$

There is no product rule and no quotient rule for integration. That absence is exactly why the rest of the chapter exists — each remaining section is a technique for handling a shape that linearity cannot touch.

3. Integration by Substitution (Textbook 7.3.1)

The reverse of the chain rule, and the most widely used technique in the chapter — Exercise 7.2 alone has 39 questions.

The method. If the integrand has the form $f(g(x))g^{\prime }(x)$, put $t=g(x)$. Then $dt=g^{\prime }(x)dx$, and the integral collapses:

$\int f(g(x))g^{\prime }(x)dx=\int f(t)dt$

What to look for. The signal is that some function appears alongside its own derivative, up to a constant multiple. In $\int 2x{e}^{x^2}dx$, the $2x$ is precisely the derivative of $x^2$, so $t=x^2$ works immediately.

Always substitute back. The answer must be expressed in the original variable $x$, not in $t$. Leaving $t$ in an indefinite-integral answer loses the final mark.

Standard results obtained this way are worth memorising, since they recur constantly:

$\int \tan xdx=-\log |\cos x|+C=\log |\sec x|+C,\int \cot xdx=\log |\sin x|+C$

$\int \sec xdx=\log |\sec x+\tan x|+C,\int \csc xdx=\log |\csc x-\cot x|+C$

The $\sec x$ result is obtained by the trick of multiplying and dividing by $(\sec x+\tan x)$, after which the numerator is exactly the derivative of the denominator.

The general pattern behind it: whenever the numerator is the derivative of the denominator, the integral is a logarithm.

$\int \frac{g^{\prime }(x)}{g(x)}dx=\log |g(x)|+C$

4. Integration Using Trigonometric Identities (Textbook 7.3.2)

Some integrands admit no substitution as they stand, but become elementary once rewritten with an identity. Exercise 7.3 is built entirely on this.

Powers of sine and cosine are reduced using the double-angle identities:

${\sin }^{2}x=\frac{1-\cos 2x}{2},{\cos }^{2}x=\frac{1+\cos 2x}{2}$

So $\int {\sin }^{2}xdx$ becomes $\frac{1}{2}\int (1-\cos 2x)dx=\frac{x}{2}-\frac{\sin 2x}{4}+C$ — a problem with no obvious substitution turned into two term-by-term integrals.

Odd powers use ${\sin }^{2}x+{\cos }^{2}x=1$ instead. For $\int {\sin }^{3}xdx$, write ${\sin }^{3}x=\sin x(1-{\cos }^{2}x)$ and substitute $t=\cos x$.

Products of different angles are split by the product-to-sum identities:

$2\sin A\cos B=\sin (A+B)+\sin (A-B),2\cos A\cos B=\cos (A+B)+\cos (A-B)$

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

A product that could not be integrated becomes a sum of two single-angle terms that can.

The triple-angle identities handle cubes directly:

${\sin }^{3}x=\frac{3\sin x-\sin 3x}{4},{\cos }^{3}x=\frac{3\cos x+\cos 3x}{4}$

The strategic point of this section is that the identity comes first and the integration is trivial afterwards. Time spent choosing the right identity is not wasted.

5. Integrals of Some Particular Functions (Textbook 7.4)

Six standard forms, each worth memorising outright. Nearly every question in Exercise 7.4 reduces to one of them.

$\int \frac{dx}{x^2-a^2}=\frac{1}{2a}\log \left|\frac{x-a}{x+a}\right|+C\int \frac{dx}{a^2-x^2}=\frac{1}{2a}\log \left|\frac{a+x}{a-x}\right|+C$

$\int \frac{dx}{x^2+a^2}=\frac{1}{a}{\tan }^{-1}\frac{x}{a}+C\int \frac{dx}{\sqrt{x^2-a^2}}=\log \left|x+\sqrt{x^2-a^2}\right|+C$

$\int \frac{dx}{\sqrt{a^2-x^2}}={\sin }^{-1}\frac{x}{a}+C\int \frac{dx}{\sqrt{x^2+a^2}}=\log \left|x+\sqrt{x^2+a^2}\right|+C$

Note the ordering carefully. The first two differ only in which way round the subtraction goes, and their answers differ correspondingly — $\frac{x-a}{x+a}$ versus $\frac{a+x}{a-x}$. Mixing them up is one of the commonest errors in the chapter.

Completing the square is the technique that makes these applicable. An integrand such as $\frac{1}{x^2+6x+13}$ is not in standard form, but:

$x^2+6x+13=(x+3{)}^2+4=(x+3{)}^2+2^2$

so with $t=x+3$ it becomes $\int \frac{dt}{t^2+2^2}=\frac{1}{2}{\tan }^{-1}\frac{t}{2}+C$.

Linear numerators. For $\int \frac{px+q}{a{x}^2+bx+c}dx$, split the numerator as $px+q=A\cdot \frac{d}{dx}(a{x}^2+bx+c)+B$. The first piece integrates to a logarithm by the $g^{\prime }/g$ rule; the second reduces to a standard form after completing the square. The same decomposition works when the quadratic sits under a square root.

6. Integration by Partial Fractions (Textbook 7.5)

The technique for rational integrands $P(x)/Q(x)$ that no substitution simplifies.

First check the degrees. The method requires a proper rational function, with $\mathrm{deg}P<\mathrm{deg}Q$. If not, divide first: $\frac{P}{Q}=(\text{polynomial})+\frac{R}{Q}$, and apply partial fractions to the remainder term only. Skipping this check is a frequent cause of an unsolvable decomposition.

Then factor $Q(x)$ and decompose according to the factor type:

Note that a repeated factor needs one term for every power up to the multiplicity, and an irreducible quadratic needs a linear numerator, not a constant. Both are routinely got wrong.

Finding the constants. Multiply through by $Q(x)$ and either compare coefficients of like powers, or substitute convenient values of $x$. Substituting the roots of the linear factors is fastest, since each choice kills all but one unknown.

After decomposing, every piece is either a logarithm, a power, or one of the standard forms from the previous section.

A disguised case worth knowing: integrands that are rational in $\sin x$ and $\cos x$ often become rational in $t$ after the substitution $t=\tan x$ or $t=\tan \frac{x}{2}$, at which point partial fractions applies.

7. Integration by Parts (Textbook 7.6)

The counterpart to the product rule, used when the integrand is a product of two unrelated functions.

$\int f(x)g(x)dx=f(x)\int g(x)dx-\int \left[f^{\prime }(x)\int g(x)dx\right]dx$

Choosing the first function. The whole difficulty is deciding which factor plays the role of $f$. The priority order is ILATE:

Whichever type comes first in this list is taken as $f$. The logic is that $f$ gets differentiated, so you want the factor whose derivative is simpler — differentiating $\log x$ gives $1/x$, a genuine simplification, whereas differentiating $e^x$ gains nothing.

The single-function trick. Integrals such as $\int \log xdx$ and $\int {\tan }^{-1}xdx$ look like they have no second factor. Write the integrand as $\log x\cdot 1$ and take $g(x)=1$; parts then works normally.

The $e^x$ special form (7.6.1). Whenever the integrand is an exponential multiplying a function plus its own derivative, the answer is immediate:

$\int e^x\left[f(x)+f^{\prime }(x)\right]dx=e^{x}f(x)+C$

Recognising this shape converts several otherwise-lengthy Exercise 7.6 questions into one line. The general version is $\int e^{ax}[f(x)+\frac{f^{\prime }(x)}{a}]dx=\frac{e^{ax}f(x)}{a}+C$.

Three special integrals (7.6.2). Derived by applying parts to the square root against $1$:

$\int \sqrt{x^2-a^2}dx=\frac{x}{2}\sqrt{x^2-a^2}-\frac{a^2}{2}\log \left|x+\sqrt{x^2-a^2}\right|+C$

$\int \sqrt{x^2+a^2}dx=\frac{x}{2}\sqrt{x^2+a^2}+\frac{a^2}{2}\log \left|x+\sqrt{x^2+a^2}\right|+C$

$\int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+C$

These are the whole content of Exercise 7.7, and the third reappears throughout Chapter 8 for areas of circles and ellipses. Watch the signs: the first has a minus before its log term, the second a plus.

8. The Definite Integral and the Fundamental Theorems (Textbook 7.7 to 7.8)

The definite integral ${\int }_a^{b}f(x)dx$ carries limits and evaluates to a number, not a family of functions. No constant of integration appears.

The area function (7.8.1). For $f\ge 0$ on $[a,b]$, define:

$A(x)={\int }_a^{x}f(t)dt$

the area under the curve from the fixed left end $a$ up to a moving right end $x$. As $x$ increases, $A(x)$ accumulates more area.

First Fundamental Theorem (7.8.2). This accumulation function is differentiable, and:

$A^{\prime }(x)=f(x)$

In words: differentiating an area function returns the original integrand. This is the statement that ties the two halves of calculus together — area and slope are inverse operations.

Second Fundamental Theorem (7.8.3). The computational form, and the one every question actually uses. If $F$ is any antiderivative of $f$ on $[a,b]$:

${\int }_a^{b}f(x)dx={\left[F(x)\right]}_a^b=F(b)-F(a)$

Why the constant does not matter. Using $F+C$ instead of $F$ gives $(F(b)+C)-(F(a)+C)$, and the constants cancel. This is why definite-integral answers never carry $C$.

The practical procedure is therefore: find any antiderivative, evaluate at the upper limit, evaluate at the lower limit, subtract in that order. Reversing the subtraction flips the sign and is a routine source of lost marks.

9. Definite Integrals by Substitution (Textbook 7.9)

Substitution works for definite integrals too, but with one change that saves considerable effort.

Change the limits with the variable. When you put $t=g(x)$, the limits must be converted as well: $x=a$ becomes $t=g(a)$, and $x=b$ becomes $t=g(b)$. Then:

${\int }_a^{b}f(g(x))g^{\prime }(x)dx={\int }_{g(a)}^{g(b)}f(t)dt$

There is then no need to substitute back. Evaluate directly in $t$ between the new limits. Students who convert back to $x$ and use the original limits usually get the right answer but waste time; those who convert back to $x$ while keeping the new limits get it wrong.

The alternative is to ignore the limits, find the indefinite integral, substitute back to $x$, and only then apply the original limits. Both routes are valid, but they must not be mixed.

Watch for a changed direction. If $g$ is decreasing, the new lower limit may exceed the new upper limit. That is fine — leave them in that order, since ${\int }_a^b=-{\int }_b^a$ handles the sign automatically.

10. Properties of Definite Integrals (Textbook 7.10)

Section 7.10 is where the chapter becomes genuinely clever. These properties routinely evaluate integrals that have no elementary antiderivative at all.

$P_0:{\int }_a^{b}f(x)dx={\int }_a^{b}f(t)dt{P}_1:{\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$

$P_2:{\int }_a^{b}f(x)dx={\int }_a^{c}f(x)dx+{\int }_c^{b}f(x)dx$

$P_3:{\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx{P}_4:{\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$

$P_6:{\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx\text{ if }f(2a-x)=f(x),=0\text{ if }f(2a-x)=-f(x)$

$P_7:{\int }_{-a}^{a}f(x)dx=2{\int }_0^{a}f(x)dx\text{if }f\text{ is even};=0\text{if }f\text{ is odd}$

$P_2$ is what handles modulus and piecewise integrands. For ${\int }_{-1}^2|x|dx$, split at $x=0$ where the definition changes, and integrate each piece with its own formula.

$P_7$ is the fastest win in the chapter. Before doing any work on a symmetric interval $[-a,a]$, test whether $f(-x)=-f(x)$. If so the answer is $0$ immediately, with no antiderivative required.

$P_3$ and $P_4$ drive the "$I=$ something $-I$" trick. Apply the substitution, add the result to the original, and the unknown integral often cancels or collapses to something elementary.

Worked, mirroring the textbook's own technique. Evaluate $I={\int }_0^{\pi /2}\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx$.

Apply $P_4$ with $x\to \frac{\pi }{2}-x$. Since $\sin \left(\frac{\pi }{2}-x\right)=\cos x$ and $\cos \left(\frac{\pi }{2}-x\right)=\sin x$, the integrand inverts:

$I={\int }_0^{\pi /2}\log \left(\frac{4+3\cos x}{4+3\sin x}\right)dx={\int }_0^{\pi /2}-\log \left(\frac{4+3\sin x}{4+3\cos x}\right)dx=-I$

So $2I=0$, giving $I=0$ — without ever finding an antiderivative, which in this case does not exist in elementary terms.

Summary

• Integration is the inverse of differentiation: $\int f(x)dx=F(x)+C$ where $F^{\prime }(x)=f(x)$; the constant is compulsory in indefinite answers.
• Integration is linear, but there is no product or quotient rule — which is why every later section is a separate technique.
• Substitution reverses the chain rule: spot a function appearing with its own derivative, put $t=g(x)$, and always substitute back at the end.
• $\int \frac{g^{\prime }(x)}{g(x)}dx=\log |g(x)|+C$ whenever the numerator is the derivative of the denominator.
• Trigonometric identities come first and the integration is trivial after: double-angle for even powers, ${\sin }^2+{\cos }^2=1$ for odd powers, product-to-sum for mixed angles.
• Six standard forms (section 7.4) turn most "particular function" integrals into a log, ${\tan }^{-1}$ or ${\sin }^{-1}$; completing the square is what makes them applicable.
• Partial fractions need a proper rational function — divide first if not. Repeated factors need a term per power; irreducible quadratics need a linear numerator.
• Integration by parts prioritises the first function by ILATE; write $\log x$ as $\log x\cdot 1$ for single-function cases.
• $\int e^x[f(x)+f^{\prime }(x)]dx=e^{x}f(x)+C$ collapses several questions to one line when recognised.
• The three square-root integrals of 7.6.2 are the whole of Exercise 7.7 and reappear for areas in Chapter 8; watch the sign on each log term.
• First Fundamental Theorem: $A^{\prime }(x)=f(x)$ for the area function. Second: ${\int }_a^{b}f=F(b)-F(a)$, which is what every question actually uses.
• For definite integrals by substitution, change the limits along with the variable and do not substitute back.
• Properties $P_1$ to $P_7$ often avoid antiderivatives entirely: $P_2$ splits modulus integrands, $P_7$ kills odd functions on symmetric intervals instantly, and $P_3$/$P_4$ drive the $I=-I$ cancellation trick.
• The "definite integral as a limit of a sum" derivation is not part of the current edition — only a passing mention survives.