Inverse Substitution

In the substitution method, we introduce $u = g (x)$ . Inverse substitution reverses this direction: we write $x$ as a function of a new variable $t$ , setting $x = ϕ (t)$ , so that the integrand simplifies dramatically.

In the substitution method, we introduce a new variable $u$ as a function of $x$ , writing $u = g (x)$ , and replace the integral in $x$ with one in $u$ . Inverse substitution reverses this direction: we write $x$ as a function of a new variable $t$ , setting $x = ϕ (t)$ , and transform the integral accordingly. If $x = ϕ (t)$ , then dx = \phi'(t)\,dt, and the integral becomes

\int f(x)\,dx = \int f\bigl(\phi(t)\bigr)\,\phi'(t)\,dt.

The substitution $x = ϕ (t)$ must be invertible on the interval of integration, so that once we evaluate the right-hand side, we can return to the original variable $x$ .

The power of inverse substitution lies in choosing $ϕ$ so that the transformed integrand simplifies dramatically. A radical that resists direct attack often collapses into a trigonometric or hyperbolic identity after a well-chosen substitution. But the technique is not limited to radicals. Any time a change of the dependent variable $x$ simplifies the structure of the integrand, inverse substitution applies.

Worked Examples

Using the substitution $x = \tan t$ , evaluate $\int \frac{d x}{1 + x^{2}}$ .

Solution

Many students recall that $\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$ , and simply write down $\arctan x + C$ . Here we derive that result from scratch using inverse substitution, so the method is clear.

Set $x = \tan t$ , so that $d x = \sec^{2} t d t$ . The denominator becomes

1 + x^{2} = 1 + \tan^{2} t = \sec^{2} t .

Substituting,

\begin{aligned} \int\frac{dx}{1+x^2} &= \int\frac{\sec^2 t}{\sec^2 t}\,dt \\ &= \int dt \\ &= t + C. \end{aligned}

Since $x = \tan t$ , we have $t = \arctan x$ , also written $\tan^{- 1} x$ , and so

\int \frac{d x}{1 + x^{2}} = \arctan x + C .

Using the substitution $x = \sin t$ , evaluate $\int \frac{d x}{\sqrt{1 - x^{2}}}$ .

Solution

One may recall that $\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}$ , and write down $\arcsin x + C$ directly. Again, we derive this using inverse substitution.

Set $x = \sin t$ with $t \in (- \frac{π}{2}, \frac{π}{2})$ , so that $d x = \cos t d t$ . On this interval $\cos t > 0$ , so

\sqrt{1 - x^{2}} = \sqrt{1 - \sin^{2} t} = | \cos t | = \cos t .

Substituting,

\begin{aligned} \int\frac{dx}{\sqrt{1-x^2}} &= \int\frac{\cos t}{\cos t}\,dt \\ &= \int dt \\ &= t + C. \end{aligned}

Since $x = \sin t$ , we have $t = \arcsin x$ (also written $\sin^{- 1} x$ ), and so

\int \frac{d x}{\sqrt{1 - x^{2}}} = \arcsin x + C .

Using the substitution $x = \tanh t$ , evaluate $\int \frac{d x}{1 - x^{2}}$ , $| x | < 1$ .

Solution

This integral can be handled by partial fractions, but it also yields cleanly to inverse substitution with a hyperbolic function, and parallels the first example closely.

Set $x = \tanh t$ , so that $d x = {sech}^{2} t d t$ . The denominator becomes

1 - x^{2} = 1 - \tanh^{2} t = {sech}^{2} t .

Substituting,

\begin{aligned} \int\frac{dx}{1-x^2} &= \int\frac{\operatorname{sech}^2 t}{\operatorname{sech}^2 t}\,dt \\ &= \int dt \\ &= t + C. \end{aligned}

Since $x = \tanh t$ , we have $t = arctanh x$ , and so

\int \frac{d x}{1 - x^{2}} = arctanh x + C .

Using the substitution $x = \sinh t$ , evaluate $\int \frac{d x}{\sqrt{1 + x^{2}}}$ .

Solution

This is the hyperbolic analogue of the second example. Set $x = \sinh t$ , so that $d x = \cosh t d t$ . Since $\cosh t > 0$ for all $t$ ,

\sqrt{1 + x^{2}} = \sqrt{1 + \sinh^{2} t} = \cosh t .

Substituting,

\begin{aligned} \int\frac{dx}{\sqrt{1+x^2}} &= \int\frac{\cosh t}{\cosh t}\,dt \\ &= \int dt \\ &= t + C. \end{aligned}

Since $x = \sinh t$ , we have $t = arcsinh x$ , and so

\int \frac{d x}{\sqrt{1 + x^{2}}} = arcsinh x + C .

Inverse Substitution

Inverse Substitution

Worked Examples

Quick Links

Social Media