Integrals

Evaluate ${\int }_{-1}^1\frac{dx}{x^2}$.

The antiderivative of $x^{-2}$ is $-x^{-1}$, so the fundamental theorem seems to give

${\left[-\frac{1}{x}\right]}_{-1}^1=(-1)-(1)=-2.$

That answer is impossible. The integrand is positive wherever it is defined, so no reasonable value of the integral can be negative. The true situation is that the integral does not exist: the area under $x^{-2}$ near the origin is unbounded on both sides.

The fundamental theorem is a conditional statement. It says that if $f$ is continuous on $[a,b]$ and $F^{\prime }=f$ there, then ${\int }_a^{b}f=F(b)-F(a)$.

Here $f$ is not continuous on $[-1,1]$ — it is not even defined at the origin — and $-x^{-1}$ is not an antiderivative on the whole interval, only on each side separately. Applying the formula across a point where the integrand blows up is the single most productive trap in the chapter, and the tell is always an answer whose sign is impossible.

1. Reaching a standard form

Almost every indefinite integral on the paper is one of a dozen standard forms after a substitution, and the skill tested is recognising which. The three moves that do most of the work are completing the square in a quadratic denominator, substituting for whatever appears inside a composite function, and dividing numerator and denominator by a power of $x$ to create a new variable.

Illustration 1

Evaluate $\int \frac{dx}{x^2+4x+13}$.

Completing the square gives $(x+2{)}^2+9$, so with $u=x+2$ this is the arctangent form:

$\int \frac{du}{u^2+9}=\frac{1}{3}{\tan }^{-1}\frac{u}{3}+C=\frac{1}{3}{\tan }^{-1}\frac{x+2}{3}+C.$

The same completion handles $\frac{1}{\sqrt{\text{quadratic}}}$, which becomes an inverse sine or a logarithm depending on the sign of the leading coefficient.

Illustration 2

Evaluate $\int \frac{x^2+1}{x^4+1}dx$.

Nothing standard is visible, so divide numerator and denominator by $x^2$:

$\int \frac{1+\frac{1}{x^2}}{x^2+\frac{1}{x^2}}dx.$

Now $t=x-\frac{1}{x}$ has $dt=\left(1+\frac{1}{x^2}\right)dx$, which is exactly the numerator, and $x^2+\frac{1}{x^2}=t^2+2$. So the integral is $\int \frac{dt}{t^2+2}=\frac{1}{\sqrt{2}}{\tan }^{-1}\frac{x^2-1}{\sqrt{2}x}+C$.

The companion integral with $x^2-1$ on top uses $t=x+\frac{1}{x}$ instead, and recognising which numerator calls for which substitution is the whole of this family.

Two further devices complete the toolkit for algebraic integrands. When a linear numerator sits over a quadratic, or over the square root of one, split it into a multiple of the derivative of the quadratic plus a constant:

$px+q=\lambda \left(2ax+b\right)+\mu .$

The first part integrates to a logarithm or a square root immediately, and the second reduces to a completed-square standard form. Matching the coefficient of $x$ gives $\lambda$ and the constant term gives $\mu$, so the split costs one line.

When a square root of a quadratic resists, a trigonometric substitution removes it outright. The three cases are $x=a\sin \theta$ for $\sqrt{a^2-x^2}$, $x=a\tan \theta$ for $\sqrt{a^2+x^2}$, and $x=a\sec \theta$ for $\sqrt{x^2-a^2}$; in each the Pythagorean identity turns the radical into a single trigonometric function. A quadratic that is not already in one of these forms is completed to a square first.

Illustration 3

Evaluate $\int \frac{2x+3}{\sqrt{x^2+4x+13}}dx$.

The derivative of the quadratic is $2x+4$, so write $2x+3=(2x+4)-1$. The first piece integrates directly to $2\sqrt{x^2+4x+13}$. The second is $-\int \frac{dx}{\sqrt{(x+2{)}^2+9}}$, a standard logarithmic form. So

$\int =2\sqrt{x^2+4x+13}-\ln \left|x+2+\sqrt{x^2+4x+13}\right|+C.$

For a rational function of $\sin x$ and $\cos x$ there is one substitution that always works, even when nothing else does. Putting $t=\tan \frac{x}{2}$ gives

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

which converts the whole integrand into a rational function of $t$, to be finished by partial fractions. It is the last resort rather than the first, because it usually produces a messier expression than a well-chosen alternative, but it never fails on that family.

2. Integration by parts, and two patterns worth knowing

The formula $\int udv=uv-\int vdu$ is only as good as the choice of $u$, and the usual ordering — inverse trigonometric, logarithmic, algebraic, trigonometric, exponential — ranks functions by how much simpler they become on differentiation.

Two patterns recur often enough to be recognised rather than derived. The first is

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

which is just the product rule read backwards. The second is the integral that reproduces itself after two applications of parts, as with $\int e^x\sin xdx$, where the original integral reappears and is solved for algebraically.

Illustration 4

Evaluate $\int e^x\frac{1+x}{(2+x{)}^2}dx$.

Split the fraction looking for $f+f^{\prime }$. Taking $f(x)=\frac{1}{2+x}$ gives $f^{\prime }(x)=-\frac{1}{(2+x{)}^2}$, so

$f+f^{\prime }=\frac{1}{2+x}-\frac{1}{(2+x{)}^2}=\frac{(2+x)-1}{(2+x{)}^2}=\frac{1+x}{(2+x{)}^2}.$

The answer is therefore $\frac{e^x}{2+x}+C$, with no integration performed at all. Spotting the pattern is worth several minutes on a question that is otherwise a long partial-fraction exercise.

Illustration 5

Evaluate $I=\int e^x\sin xdx$.

Two applications of parts give $I=e^x\sin x-e^x\cos x-I$, so $2I=e^x\left(\sin x-\cos x\right)$ and

$I=\frac{e^x}{2}\left(\sin x-\cos x\right)+C.$

The essential point is to differentiate the same factor both times. Swapping roles on the second application returns the trivial identity $I=I$ and wastes the work.

3. Rational functions

A rational function is integrated by reducing the degree of the numerator below that of the denominator, factorising the denominator, and splitting into partial fractions. A linear factor $(x-a)$ contributes $\frac{A}{x-a}$, a repeated factor $(x-a{)}^2$ contributes an extra $\frac{B}{(x-a{)}^2}$, and an irreducible quadratic contributes $\frac{Cx+D}{x^2+bx+c}$.

Illustration 6

Evaluate $\int \frac{x^2}{(x-1)(x^2+1)}dx$.

Write $\frac{x^2}{(x-1)(x^2+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}$. Multiplying out and setting $x=1$ gives $A=\frac{1}{2}$; comparing coefficients of $x^2$ and of the constant gives $B=\frac{1}{2}$ and $C=\frac{1}{2}$. So

$\int =\frac{1}{2}\ln |x-1|+\frac{1}{4}\ln \left(x^2+1\right)+\frac{1}{2}{\tan }^{-1}x+C.$

Substituting a root to find one coefficient and then comparing two coefficients is faster than expanding everything, and it leaves fewer places to slip.

4. Definite integrals and their symmetries

A definite integral has properties an indefinite one does not, all of them consequences of substituting a reflection or a shift. The most productive is the king property:

${\int }_a^{b}f(x)dx={\int }_a^{b}f(a+b-x)dx,$

obtained by substituting $x\mapsto a+b-x$. Adding the two forms often collapses an intractable integrand into a constant.

Alongside it, an odd integrand over a symmetric interval gives zero and an even one gives twice the half-integral; a function of period $T$ satisfies ${\int }_0^{nT}f=n{\int }_0^{T}f$; and ${\int }_0^{2a}f(x)dx=2{\int }_0^{a}f(x)dx$ when $f(2a-x)=f(x)$, and zero when $f(2a-x)=-f(x)$.

Illustration 7

Show that ${\int }_0^{\pi /2}\frac{dx}{1+{\tan }^{n}x}=\frac{\pi }{4}$ for every real $n$.

Call the integral $I$ and apply the king property with $a+b=\frac{\pi }{2}$. Since $\tan \left(\frac{\pi }{2}-x\right)=\cot x$,

$I={\int }_0^{\pi /2}\frac{dx}{1+{\cot }^{n}x}={\int }_0^{\pi /2}\frac{{\tan }^{n}x}{1+{\tan }^{n}x}dx.$

Adding the two expressions, the integrands sum to $1$, so $2I=\frac{\pi }{2}$ and $I=\frac{\pi }{4}$. The exponent never appears in the answer, which is the surprise the question is built around.

Illustration 8

Evaluate ${\int }_0^{\pi }\frac{x\sin x}{1+{\cos }^{2}x}dx$.

Substituting $x\mapsto \pi -x$ leaves $\sin x$ and ${\cos }^{2}x$ unchanged but turns $x$ into $\pi -x$. Adding the two forms,

$2I=\pi {\int }_0^{\pi }\frac{\sin x}{1+{\cos }^{2}x}dx=\pi {\left[-{\tan }^{-1}(\cos x)\right]}_0^{\pi }=\pi \cdot \frac{\pi }{2},$

so $I=\frac{{\pi }^2}{4}$. The general pattern is that ${\int }_0^{\pi }xf(\sin x)dx=\frac{\pi }{2}{\int }_0^{\pi }f(\sin x)dx$, and it is worth recognising rather than rederiving each time.

One more property is used whenever an integral cannot be evaluated at all. If $f\le g$ throughout $[a,b]$ then ${\int }_a^{b}f\le {\int }_a^{b}g$, and in particular a bound $m\le f\le M$ gives

$m(b-a)\le {\int }_a^{b}f(x)dx\le M(b-a).$

Advanced sets questions asking to place an integral between two numbers, and the intended method is always to bound the integrand rather than to find an antiderivative. To estimate ${\int }_0^1\frac{x^2}{1+x^4}dx$, note that $1\le 1+x^4\le 2$ on $[0,1]$, so the integrand lies between $\frac{x^2}{2}$ and $x^2$. Integrating the two bounds gives

$\frac{1}{6}\le {\int }_0^1\frac{x^2}{1+x^4}dx\le \frac{1}{3},$

and a sharper bound comes from a sharper estimate of the denominator, not from more integration. The same idea shows that an integral of a strictly positive function over an interval of positive length is strictly positive, which is the sanity check that exposed the opening example.

5. A definite integral as a limit of a sum

Dividing $[0,1]$ into $n$ equal parts and letting $n\to \infty$,

${\lim }_{n\to \infty }\frac{1}{n}{\sum }_{r=1}^{n}f\left(\frac{r}{n}\right)={\int }_0^{1}f(x)dx.$

Reading a limit as an integral requires spotting the factor $\frac{1}{n}$ and rewriting the rest as a function of $\frac{r}{n}$ alone. Advanced sets this in the direction that is harder: a sum is given and the integral must be found.

Illustration 9

Evaluate $\underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{n}{n^2+r^2}$.

Factor $n^2$ out of the denominator to expose the pattern:

${\sum }_{r=1}^n\frac{n}{n^2+r^2}=\frac{1}{n}{\sum }_{r=1}^n\frac{1}{1+{\left(\frac{r}{n}\right)}^2}⟶{\int }_0^1\frac{dx}{1+x^2}=\frac{\pi }{4}.$

The first step is always to force the leading $\frac{1}{n}$ into view; everything else then has to be a function of $\frac{r}{n}$, and if it is not, the sum is not a Riemann sum on $[0,1]$.

The same reading handles a sum written out in full. In $\frac{1}{n}\left[\sqrt{\frac{1}{n}}+\sqrt{\frac{2}{n}}+\cdots +\sqrt{\frac{n}{n}}\right]$ every term is $\sqrt{r/n}$ with the prefactor already present, so the limit is ${\int }_0^1\sqrt{x}dx=\frac{2}{3}$.

6. Differentiating an integral

When the limits themselves depend on $x$, the derivative is given by Leibniz's rule:

$\frac{d}{dx}{\int }_{g(x)}^{h(x)}f(t)dt=f\left(h(x)\right)h^{\prime }(x)-f\left(g(x)\right)g^{\prime }(x).$

The integration variable is $t$, not $x$, which is why the answer contains no integral at all. Advanced uses this to define a function by an integral and then ask for its monotonicity or extrema.

Illustration 10

If $F(x)={\int }_x^{x^2}\frac{dt}{\ln t}$ for $x>1$, find $F^{\prime }(x)$.

Applying the rule with $h(x)=x^2$ and $g(x)=x$,

$F^{\prime }(x)=\frac{2x}{\ln x^2}-\frac{1}{\ln x}=\frac{2x}{2\ln x}-\frac{1}{\ln x}=\frac{x-1}{\ln x}.$

Since $x>1$ makes both parts positive, $F$ is increasing — a conclusion reached without ever evaluating the integral, which in fact has no elementary form.

7. Areas, and why the region must be split

The area between two curves is $\int \left(\text{upper}-\text{lower}\right)dx$, and the words "upper" and "lower" can swap at a crossing. An integral run straight through such a point subtracts one piece from the other instead of adding them.

Illustration 11

Find the area enclosed between $y=x$ and $y=x^3$ for $-1\le x\le 1$.

The curves meet at $x=-1,0,1$. On $(0,1)$ the line is above the cubic; on $(-1,0)$ the cubic is above the line. So

$A={\int }_{-1}^0\left(x^3-x\right)dx+{\int }_0^1\left(x-x^3\right)dx=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$

Running a single integral of $x-x^3$ from $-1$ to $1$ gives $0$, because the integrand is odd. A zero answer to an area question is always a signal that a crossing was ignored.

Illustration 12

Find the area bounded by $y^2=4x$ and the line $y=2x-4$.

Integrating with respect to $y$ avoids splitting: the parabola gives $x=\frac{y^2}{4}$ and the line gives $x=\frac{y+4}{2}$, meeting where $y^2=2y+8$, that is $y=-2$ and $y=4$. Then

$A={\int }_{-2}^4\left(\frac{y+4}{2}-\frac{y^2}{4}\right)dy={\left[\frac{y^2}{4}+2y-\frac{y^3}{12}\right]}_{-2}^4=9.$

Choosing the variable of integration so that one curve is a single function of it is usually worth more than any algebraic simplification.

8. Reduction and the standard trigonometric integrals

Integrals of ${\sin }^{n}x$ over a quarter period satisfy a reduction formula, and iterating it gives Wallis's result:

${\int }_0^{\pi /2}{\sin }^{n}xdx=\frac{n-1}{n}{\int }_0^{\pi /2}{\sin }^{n-2}xdx,$

with the chain terminating at ${\int }_0^{\pi /2}dx=\frac{\pi }{2}$ for even $n$ and at ${\int }_0^{\pi /2}\sin xdx=1$ for odd $n$. The same value holds with ${\cos }^{n}x$, by the king property.

Illustration 13

Evaluate ${\int }_0^{\pi /2}{\sin }^{6}xdx$ and ${\int }_0^{\pi /2}{\sin }^{5}xdx$.

For the even case, $\frac{5}{6}\cdot \frac{3}{4}\cdot \frac{1}{2}\cdot \frac{\pi }{2}=\frac{15\pi }{96}=\frac{5\pi }{32}$.

For the odd case, $\frac{4}{5}\cdot \frac{2}{3}\cdot 1=\frac{8}{15}$.

The factor of $\frac{\pi }{2}$ appears only for even powers, which is the one detail to keep straight.

Illustration 14

Evaluate ${\int }_0^{2\pi }\left|\sin x\right|dx$.

The integrand has period $\pi$, not $2\pi$, so the integral is $2{\int }_0^{\pi }\sin xdx=2\times 2=4$. Dropping the modulus and integrating $\sin x$ over $[0,2\pi ]$ gives $0$, which is the same failure as the area question above: a sign change inside the interval.

Illustration 15

Evaluate ${\int }_{-2}^2\frac{x^2{\tan }^{-1}x}{1+x^4}dx$.

Test the parity. The factor $\frac{x^2}{1+x^4}$ is even and ${\tan }^{-1}x$ is odd, so the product is odd and the integral over a symmetric interval is $0$. No antiderivative exists in elementary terms, so parity is not merely the quick route — it is the only one.

Summary

The fundamental theorem needs the integrand continuous on the whole interval. Applying an antiderivative across a point where the function blows up produces an answer that is often impossible on sight, as with ${\int }_{-1}^1{x}^{-2}dx$ appearing to be negative.

The corresponding check is worth running on every definite integral before the answer is written down: compare the sign of the result with the sign of the integrand, and compare its size with the crude bound given by the length of the interval.

Indefinite integrals are exercises in recognition: complete the square for a quadratic denominator, substitute for whatever sits inside a composite, and divide through by a power of $x$ to create $x\pm \frac{1}{x}$. In integration by parts, choose the factor that simplifies on differentiation, and recognise the two standard patterns — the product rule read backwards for $e^x\left[f+f^{\prime }\right]$, and the integral that reappears after two applications.

A linear numerator over a quadratic, or over its square root, is split into a multiple of the derivative of that quadratic plus a constant. A surviving square root is removed by a sine, tangent or secant substitution, and any rational function of $\sin x$ and $\cos x$ yields to $t=\tan \frac{x}{2}$, which is slow but never fails.

Definite integrals carry symmetries that indefinite ones do not. The king property replaces $x$ by $a+b-x$ and, added to the original, often collapses the integrand to a constant; parity kills an odd integrand over a symmetric interval; and periodicity converts a long interval into a multiple of one period.

When an integral cannot be evaluated, it can still be bounded: replace the integrand by constants above and below it and multiply by the length of the interval. That is the intended route whenever a question asks you to place an integral between two numbers, and the same reasoning is what makes an impossible sign impossible.

A limit of a sum becomes an integral once the factor $\frac{1}{n}$ is visible and everything else is a function of $\frac{r}{n}$. Leibniz's rule differentiates an integral with variable limits without evaluating it, which is how a function defined by an integral is analysed. For areas, find every crossing first and integrate each piece with the correct order of subtraction; an answer of zero means a crossing was missed.