Integrals

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 7.

Can you rebuild a journey from the speedometer?

Differentiation takes a position and gives you the speed. Integration goes the other way: given the speed at every instant, it recovers the position. Notice the catch. Knowing only how fast you were moving never tells you where you started, so the recovered position is only fixed up to a starting point. That one idea explains the mysterious "$+\,C$" in every indefinite integral, and it is the heart of this chapter.

The chapter in one line: antiderivative → standard forms → four methods (substitution, special forms, partial fractions, parts) → definite integral → properties that save time.

Antiderivatives and the constant $C$

A function $F$ is an antiderivative of $f$ if $F'(x) = f(x)$. The set of all antiderivatives is the indefinite integral:

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

Here $f(x)$ is the integrand, $x$ is the variable of integration and $C$ is the constant of integration.

Why exactly $+\,C$? If two functions have the same derivative on an interval, they differ by a constant, because a function whose derivative is zero is constant. So once you find one antiderivative $F$, every other one is $F + C$. Geometrically, $F + C$ is a family of curves, all copies of one curve slid up or down, and they have the same slope at each $x$.

Integration obeys two simple rules that let you work term by term:

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

Also, differentiating an integral gives back the integrand, $\dfrac{d}{dx}\int f(x)\,dx = f(x)$, and $\int f'(x)\,dx = f(x) + C$.

The standard integrals come straight from the derivatives you already know, read backwards.

Why $\dfrac{1}{x}$ gives a logarithm. The power rule would ask for $\dfrac{x^0}{0}$, which is meaningless. The gap is filled by $\log\lvert x\rvert$, whose derivative is $\dfrac{1}{x}$ for both positive and negative $x$.

Beyond the textbook. For the exam: any correct antiderivative plus $C$ is a full answer. In reality: two answers that look different can both be right. For instance $\int \sin x\cos x\,dx$ can be written as $\dfrac{\sin^2 x}{2}$ or as $-\dfrac{\cos^2 x}{2}$, and these differ by the constant $\dfrac{1}{2}$, which $C$ absorbs. Why it matters: when your answer is not among the options, check whether an option differs from yours by a constant before you decide you are wrong.

Method 1: substitution

The idea. The chain rule says $\dfrac{d}{dx}F(g(x)) = F'(g(x))\,g'(x)$. Read backwards, an integrand of the shape "a function of $g(x)$ times $g'(x)$" integrates easily. In practice you put $t = g(x)$, so $dt = g'(x)\,dx$, and the integral becomes one in $t$.

Look for a function and its derivative appearing together. One pattern is so common that it deserves its own line:

$$\int \dfrac{f'(x)}{f(x)}\,dx = \log\lvert f(x)\rvert + C$$

It produces the integrals of the four remaining trigonometric functions.

When the integrand has powers of sine and cosine, an identity usually reduces it to standard forms first.

Worked example: a function and its derivative. Find $\int x\sqrt{x^2 + 1}\,dx$.

Common wrong step. Writing $\int t^{1/2}\,dt$ and forgetting the $\dfrac{1}{2}$ that comes from $x\,dx = \dfrac{dt}{2}$. After any substitution, check by differentiating your answer.

Pause and predict: without calculating, what is $\int \cos 5x\,dx$?

Check your answer: $\dfrac{\sin 5x}{5} + C$. Differentiating $\sin 5x$ produces an extra factor $5$ from the chain rule, so the integral must carry $\dfrac{1}{5}$ to cancel it. A linear inside $ax + b$ always brings $\dfrac{1}{a}$.

Method 2: the six special forms

Six integrals appear so often that you should know them as formulas. All have $a > 0$.

Where these come from. The first row follows from the split $\dfrac{1}{x^2 - a^2} = \dfrac{1}{2a}\left( \dfrac{1}{x - a} - \dfrac{1}{x + a} \right)$, which integrates to two logarithms and combines into one. The $\tan^{-1}$ row comes from the substitution $x = a\tan\theta$, and the $\sin^{-1}$ row from $x = a\sin\theta$.

How to remember the pattern. A sum $x^2 + a^2$ in the denominator gives an inverse tangent with the factor $\dfrac{1}{a}$. A difference of squares gives a logarithm with $\dfrac{1}{2a}$. A square root with $a^2 - x^2$ gives an inverse sine, and the two other roots give logarithms.

Quadratics that are not in standard form. For $\int \dfrac{dx}{ax^2 + bx + c}$ or $\int \dfrac{dx}{\sqrt{ax^2 + bx + c}}$, complete the square first. Writing

$$ax^2 + bx + c = a\left[ \left( x + \dfrac{b}{2a} \right)^2 + \left( \dfrac{c}{a} - \dfrac{b^2}{4a^2} \right) \right]$$

turns it into one of the six forms with $x$ replaced by $x + \dfrac{b}{2a}$. When the numerator is $px + q$, split it as $px + q = A\,\dfrac{d}{dx}\left( ax^2 + bx + c \right) + B$. The first piece integrates to a logarithm (or a root) and the second is a form you already know.

Worked example: completing the square. Find $\int \dfrac{dx}{x^2 + 2x + 5}$.

Common wrong step. Reaching for a logarithm because the denominator is a quadratic. Check the sign after completing the square: a plus between the squares means $\tan^{-1}$, a minus means a logarithm.

Method 3: partial fractions

A rational function is a ratio of two polynomials. If it is hard to integrate as a whole, split it into simple pieces that each integrate to a logarithm or a power. This works for a proper fraction, where the degree of the numerator is less than the degree of the denominator. If the fraction is improper, divide first, so that $\dfrac{P(x)}{Q(x)} = T(x) + \dfrac{P_1(x)}{Q(x)}$. Then $T(x)$ is a polynomial and $\dfrac{P_1(x)}{Q(x)}$ is proper.

The form of the split depends on the denominator.

Why a repeated factor needs two terms. For $(x - a)^2$, a single term $\dfrac{A}{(x - a)^2}$ cannot combine into every possible numerator. You need both powers. Why an irreducible quadratic needs $Bx + D$. The numerator must be one degree lower than the quadratic, so a linear numerator is the general choice.

Worked example: two linear factors. Find $\int \dfrac{3x + 5}{(x - 1)(x + 3)}\,dx$.

Common wrong step. Splitting an improper fraction such as $\dfrac{x^2 + 1}{(x - 1)(x + 2)}$ directly. The numerator degree equals the denominator degree, so you must divide first.

Method 4: integration by parts

Where it comes from. It is the product rule read backwards. For two functions $u$ and $w$, $\dfrac{d}{dx}(uw) = u'w + uw'$. Integrate both sides and rearrange. If $w$ is an antiderivative of $v$, you get:

$$\int u\,v\,dx = u\int v\,dx - \int \left( \dfrac{du}{dx}\int v\,dx \right) dx$$

In words: the integral of a product is (first) × (integral of second) − the integral of (derivative of first × integral of second).

Which function goes first? You want the first function to become simpler when differentiated and the second to be easy to integrate. The order I L A T E (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential) says which comes first when two types meet. The earlier letter is taken as the first function.

For a single function like $\log x$ or $\tan^{-1} x$, write it as that function times $1$ and take the function as first.

Worked example: choosing the first function. Find $\int x\cos x\,dx$.

Common wrong step. Taking $\cos x$ as first. Then you get $\cos x\cdot\dfrac{x^2}{2} + \int \dfrac{x^2}{2}\sin x\,dx$, and the new integral is worse, because the power of $x$ went up. If the second integral looks harder than the first, swap the roles.

Worked example: log x on its own. Find $\int \log x\,dx$.

Pause and predict: which function is the first function in $\int x^2\log x\,dx$, and what is the integral?

Check your answer: $\log x$ is first, by ILATE. Then $\int x^2\log x\,dx = \dfrac{x^3}{3}\log x - \int \dfrac{x^3}{3}\cdot\dfrac{1}{x}\,dx = \dfrac{x^3}{3}\log x - \dfrac{x^3}{9} + C$.

A shortcut worth knowing. If the integrand is $e^x\left[ f(x) + f'(x) \right]$, the answer is simply $e^x f(x) + C$, because the by-parts terms cancel. For example, $\int e^x\left( \sin x + \cos x \right) dx = e^x\sin x + C$.

Three root integrals. Integration by parts also gives these, and the third is the one behind the area of a circle in the next chapter.

Beyond the textbook. For the exam: every integral in the paper has an answer you can write using the functions in this chapter. In reality: many innocent looking integrals do not. For example $\int e^{x^2}\,dx$ cannot be written with polynomials, exponentials, logarithms, trigonometric functions or their inverses, no matter how clever the method. Why it matters: such integrals are handled by numerical methods on computers, which is how areas and probabilities that have no neat formula are actually found.

The definite integral

A definite integral $\int_a^b f(x)\,dx$ is a single number, the signed area between the graph of $f$, the $x$-axis and the lines $x = a$ and $x = b$. Think of the area function $A(x) = \int_a^x f(t)\,dt$, the area collected as you sweep from $a$ to $x$. When $x$ moves by a tiny amount $h$, the area grows by about $f(x)\,h$, a thin strip. That is the content of the first fundamental theorem: $A'(x) = f(x)$, so $A$ is an antiderivative of $f$.

The second fundamental theorem turns this into a calculation. If $F$ is any antiderivative of the continuous function $f$, then:

$$\int_a^b f(x)\,dx = \left[ F(x) \right]_a^b = F(b) - F(a)$$

Why no $C$ here. An antiderivative $F + C$ gives $(F(b) + C) - (F(a) + C)$, and the $C$ cancels. A definite integral therefore has no constant. This is also why you never add $C$ to a definite integral.

Substitution in a definite integral. You may either find the indefinite integral and then use the original limits, or change the limits together with the variable and never return to $x$. The second way is faster, but the limits must change.

Worked example: change the limits. Evaluate $\int_0^{\pi/2} \sin x\cos^2 x\,dx$.

Common wrong step. Keeping the old limits $0$ and $\dfrac{\pi}{2}$ after switching to $t$. That integrates $t^2$ over the wrong range, $0$ to $\dfrac{\pi}{2}$, and produces an answer containing $\pi^3$, which is wrong.

Properties of definite integrals

These properties often turn a long calculation into a one-liner. Pick the property from the shape of the problem.

Why the even and odd rule holds. An even function has a graph that mirrors across the $y$-axis, so the left half of the area equals the right half. An odd function has equal and opposite halves, so one cancels the other.

Why the king property holds. Replacing $x$ by $a - x$ runs through the same values of $x$ in the opposite order. The graph is read backwards, but the area under it does not change.

Worked example: the king property at work. Evaluate $I = \int_0^{\pi/2} \dfrac{\sin x}{\sin x + \cos x}\,dx$.

Common wrong step. Writing the integral of a quotient as a quotient of integrals, $\dfrac{\int \sin x\,dx}{\int(\sin x + \cos x)\,dx}$. Integration is not division-friendly: there is no such rule.

Pause and predict: is $\int_{-1}^{1} x^2\sin x\,dx$ zero?

Check your answer: yes. $x^2$ is even and $\sin x$ is odd, so the product is odd, and an odd function over $[-1, 1]$ integrates to $0$. By contrast, $\int_{-1}^{1} x^2\,dx = \dfrac{2}{3}$, because $x^2$ is even.

Beyond the textbook. For the exam: the king property is a trick for integrals on $[0, a]$ where $f(x)$ and $f(a - x)$ combine neatly. In reality: it is a statement about symmetry. The substitution $x \to a - x$ is a reflection of the interval, and a reflection never changes the total area. The same idea explains why the integral of $\sin^n x$ and $\cos^n x$ over $\left[ 0, \dfrac{\pi}{2} \right]$ are equal. Why it matters: when the integrand has a partner under reflection (like sine and cosine), averaging the two forms removes the difficult part.

Which method first?

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Previous: Application of Derivatives uses derivatives to study rates and extremes. This chapter runs the same machinery in reverse.
• Next: Application of Integrals uses the definite integral to find areas under curves, including the circle and the ellipse.
• Earlier ideas reused: Continuity and Differentiability gives the chain and product rules behind substitution and by parts, and Inverse Trigonometric Functions gives the $\sin^{-1}$ and $\tan^{-1}$ results.
• Used later: Differential Equations solves equations by integrating both sides.