How a Clever Substitution Turns a Hard Integral Into an Easy One

Most integrands are not on the list of standard forms, but many are standard forms in disguise — a function sitting next to its own derivative. Substituting a new variable removes the disguise and reduces the integral to one you already know.

This lesson covers the substitution method, integrals of the forms $f^{\prime }(x)[f(x){]}^n$ and $\frac{f^{\prime }(x)}{f(x)}$, and integrals of tan, cot, sec and cosec.

**To integrate by substitution, choose $t=g(x)$ so that $dt=g^{\prime }(x)dx$ appears in the integrand, rewrite the whole integral in terms of t, integrate, and then substitute back for x.

Choosing t. Look for an expression whose derivative, up to a constant factor, also appears — often the inside of a bracket, a root or an exponent.

Worked example.** Find $\int 2x\cos (x^2)dx$. Let $t=x^2$, so $dt=2xdx$:

$$\int \cos tdt=\sin t+C=\sin (x^2)+C$$

Worked example 2. Find $\int x\sqrt{x^2+9}dx$. Let $t=x^2+9$, so $xdx=\frac{dt}{2}$:

$$\frac{1}{2}\int t^{1/2}dt=\frac{1}{3}{t}^{3/2}+C=\frac{1}{3}(x^2+9{)}^{3/2}+C$$

Check: differentiating gives $\frac{1}{3}\times \frac{3}{2}(x^2+9{)}^{1/2}\times 2x=x\sqrt{x^2+9}$.

Worked example 3. $\int e^{\sin x}\cos xdx=e^{\sin x}+C$, using $t=\sin x$.

An everyday example. Converting a recipe from cups to grams before scaling it up changes the variable to make the arithmetic easy — substitution does the same for integrals.

The substance. Every x must be replaced, including dx — an integral with both x and t left in it cannot be finished.

**When a function appears with its own derivative, $\int f^{\prime }(x)[f(x){]}^{n}dx=\frac{[f(x){]}^{n+1}}{n+1}+C$ for $n\ne -1$, and $\int \frac{f^{\prime }(x)}{f(x)}dx=\log |f(x)|+C$.

Why they work.** Put $t=f(x)$, so $dt=f^{\prime }(x)dx$; the integrals become $\int t^{n}dt$ and $\int \frac{dt}{t}$.

Worked example (power form).

$$\int {\sin }^{4}x\cos xdx=\frac{{\sin }^{5}x}{5}+C$$

Worked example (power form 2). $\int \frac{(\log x{)}^2}{x}dx=\frac{(\log x{)}^3}{3}+C$, since $\frac{1}{x}$ is the derivative of $\log x$.

Worked example (log form).

$$\int \frac{2x+3}{x^2+3x+5}dx=\log |x^2+3x+5|+C$$

Adjusting a constant. $\int \frac{x}{x^2+1}dx=\frac{1}{2}\int \frac{2x}{x^2+1}dx=\frac{1}{2}\log (x^2+1)+C$.

An everyday example. Spotting that a bill already shows its tax as a separate line saves recalculating it — spotting $f^{\prime }(x)$ beside $f(x)$ saves a full substitution.

The substance. Only constant factors can be adjusted — $\int \frac{x^2}{x^2+1}dx$ is not a log form, because $x^2$ is not a constant multiple of $2x$.

**Writing each function as a quotient reveals a derivative-over-function form: $\int \tan xdx=\log |\sec x|+C$, $\int \cot xdx=\log |\sin x|+C$, $\int \sec xdx=\log |\sec x+\tan x|+C$ and $\int \text{cosec}xdx=\log |\text{cosec}x-\cot x|+C$.

Derivation for $\tan x$.**

$$\int \frac{\sin x}{\cos x}dx=-\log |\cos x|+C=\log |\sec x|+C$$

**Derivation for $\sec x$.** Multiply above and below by $\sec x+\tan x$:

$$\int \frac{{\sec }^{2}x+\sec x\tan x}{\sec x+\tan x}dx=\log |\sec x+\tan x|+C$$

because the numerator is the derivative of the denominator.

Worked example. $\int \tan 3xdx=\frac{1}{3}\log |\sec 3x|+C$.

Worked example 2. $\int (\tan x+\cot x)dx=\log |\sec x|+\log |\sin x|+C=\log |\tan x|+C$.

Numerical check. The derivative of $\log (\sec x+\tan x)$ at $x=0.5$ equals $\sec 0.5\approx 1.139$.

An everyday example. Navigation charts on which a constant compass course is a straight line stretch distances north and south according to the integral of sec of the latitude.

The substance. The modulus matters — $\log (\cos x)$ is undefined where $\cos x<0$, but $\log |\sec x|$ works on every interval where tan is defined.