7.1 Antiderivatives and Integration by Substitution

A delivery drone in Bengaluru logs its vertical speed every second, but its altimeter has failed. The speed log says $v(t)=8-2t$ metres per second. Since speed is the rate of change of height, $h'(t)=8-2t$, and we want $h(t)$ itself. You can probably guess one answer: $h(t)=8t-t^2$ has exactly this derivative. But so does $8t-t^2+5$, and $8t-t^2-17$ — the speed log cannot tell whether the drone took off from the ground or from a 5-metre rooftop. One extra fact, such as the starting height, picks out the right function.

That is the whole of this section in miniature. Finding a function from its derivative is called integration; the answer is a family of functions differing by constants; and the family is pinned down by one known value. Because integration is differentiation read backwards, every derivative you learnt in Chapter 5 gives an integral for free. When the integrand is a chain-rule derivative in disguise, the method of substitution removes the disguise.

Integration as the Inverse of Differentiation

For example, $x^3$ is an antiderivative of $3x^2$ on $\mathbb R$, because $\dfrac{d}{dx}\left(x^3\right)=3x^2$. So are $x^3+5$, $x^3-\pi$ and $x^3+C$ for every real constant $C$, since the derivative of a constant is $0$. Are there any others? The next theorem says no — on an interval, adding constants is the only freedom.

Why. Put $H=G-F$. Then $H'(x)=G'(x)-F'(x)=f(x)-f(x)=0$ for every $x$ in $I$. A function whose derivative is zero at every point of an interval has a horizontal tangent everywhere, so its graph never rises or falls: $H$ is constant on $I$. (We accept this geometrically clear fact here; its formal proof belongs to a later course.) Hence $G-F=C$.

The word interval matters. The function $\dfrac1x$ is defined on two separate intervals, $x\lt0$ and $x\gt0$, and there an antiderivative may use one constant on the left and a different one on the right. In this course we always work on one interval at a time and write a single $C$.

So $\displaystyle\int 3x^2\,dx=x^3+C$ and, for the drone, $\displaystyle\int(8-2t)\,dt=8t-t^2+C$. Geometrically the indefinite integral is a family of curves $y=F(x)+C$, each a vertical shift of the others; at every $x$ all of them have the same slope $f(x)$ (the chapter teaser shows this). A particular antiderivative is one member of the family, chosen by a condition such as “the curve passes through $(1,3)$”.

Doing this systematically — taking each derivative formula you know and reading it from right to left — gives a list of integrals that you should know as well as your multiplication tables. We split it into three groups.

Why $\log|x|$? For $x\gt0$, $\dfrac{d}{dx}\log x=\dfrac1x$. For $x\lt0$, $\log(-x)$ is defined and, by the chain rule, $\dfrac{d}{dx}\log(-x)=\dfrac{1}{-x}\cdot(-1)=\dfrac1x$. Both cases are covered by $\log|x|$. Notice also that the power rule cannot give $\int x^{-1}\,dx$: it would divide by $n+1=0$. The logarithm fills exactly that gap. (Recall: $\log$ means $\log_e$ throughout.)

Because $\sin^{-1}x+\cos^{-1}x=\tfrac\pi2$, the answer $-\cos^{-1}x+C$ is the same family as $\sin^{-1}x+C$ — only the constant differs. Likewise $-\cot^{-1}x$ and $-\operatorname{cosec}^{-1}x$ can replace $\tan^{-1}x$ and $\sec^{-1}x$. Two correct answers to an integral may look different and still differ only by a constant.

Two rules let us integrate sums and multiples term by term; the first also says that differentiation and integration undo each other.

Why. Each is a derivative rule read backwards. If $F'=f$ and $G'=g$, then $(F\pm G)'=f\pm g$ and $(kF)'=kf$, so $F\pm G$ is an antiderivative of $f\pm g$ and $kF$ of $kf$. (The two arbitrary constants of $\int f$ and $\int g$ combine into one, which is why a single $+\,C$ is written at the end.) Together the sum and multiple rules say that integration is linear, exactly like differentiation.

Integration by Substitution

Look at $\displaystyle\int2x\cos(x^2)\,dx$. It is not in the table. But the chain rule gives $\dfrac{d}{dx}\sin(x^2)=\cos(x^2)\cdot2x$ — exactly the integrand. So the answer is $\sin(x^2)+C$. The integrand was a chain-rule derivative in disguise: an outer function ($\cos$) evaluated at an inner function ($x^2$), multiplied by the derivative of the inner function ($2x$). Substitution is the systematic way to spot and remove such a disguise.

Why. By the chain rule, $\dfrac{d}{dx}F(g(x))=F'(g(x))\,g'(x)=f(g(x))\,g'(x)$. So $F(g(x))$ is an antiderivative of the integrand, and Theorem 7.1 supplies the $+\,C$. The symbol $du=g'(x)\,dx$ is a bookkeeping device: it records that a small change $dx$ in $x$ produces a change $g'(x)\,dx$ in $u$.

A special case is used constantly. If $u=ax+b$ with $a\ne0$, then $du=a\,dx$, and $$\int f(ax+b)\,dx=\frac1a\,F(ax+b)+C.$$ For example $\displaystyle\int(3x-2)^5dx=\frac{(3x-2)^6}{18}+C$ and $\displaystyle\int\sec^2(4x)\,dx=\tfrac14\tan4x+C$.

One substitution is worth stating once and for all. If the numerator is the derivative of the denominator, put $u=f(x)$, $du=f'(x)\,dx$: then $\displaystyle\int\frac{f'(x)}{f(x)}\,dx=\int\frac{du}{u}=\log|u|+C$.

Why. $\tan x=\dfrac{\sin x}{\cos x}=-\dfrac{(\cos x)'}{\cos x}$, so $\displaystyle\int\tan x\,dx=-\log|\cos x|+C$, which is $\log|\sec x|+C$. For $\sec x$, multiply above and below by $\sec x+\tan x$: $$\sec x=\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}$$ and the numerator is exactly the derivative of the denominator. The cotangent and cosecant results are proved the same way (Exercise III).

Integrals Using Trigonometric Identities

Many trigonometric integrands are not in the table but can be rewritten as sums of things that are. Three families of identities do almost all the work:

For an odd power of $\sin x$ or $\cos x$ there is a second route: split off one factor and use $\sin^2x+\cos^2x=1$ so that the rest is a function of the other one, then substitute.

Exercises