Integration by Substitution

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How substitution simplifies integration

Integration by substitution is a technique used to simplify an integral by introducing a suitable substitution. When the integral is not straightforward to compute, this method proves highly useful as it allows rewriting the integral of a function $f(x)$ in terms of a new variable $u$, simplifying the computation:

\int f(g(x))\,g'(x)\,dx = \int f(u)\,du

The process involves the following steps:

  • Introduce a change of variable by defining $u = g(x)$, where $g(x)$ is an appropriately chosen function.
  • Compute the differential transformation, given by $du = g'(x)dx.$
  • Rewrite the integral in terms of $u$, replacing $x$ and $dx$ accordingly, to obtain an equivalent expression that is often more straightforward to solve.
  • Once the integral is evaluated, revert to the original variable $x$ to express the final result in its initial form.

The key insight is that substitution reverses the chain rule: recognizing this connection makes it easier to identify when and how to apply the technique.

The method of substitution is a direct consequence of the chain rule for derivatives. If $F(x) = H(g(x))$, then by the chain rule:

F'(x) = H'(g(x))\, g'(x)

Therefore, whenever an integrand has the form $H'(g(x))\, g'(x)$ it is the derivative of the composite function $H(g(x))$. Integration by substitution simply reverses this process by introducing $u = g(x)$, reducing the integral to:

\int H'(u)\, du = H(u) + c

Recognizing when to use substitution

Before proceeding to concrete examples, it is useful to understand when a substitution is likely to be effective. The technique is most natural when the integrand contains a composite function. In many cases, the integral has the general form:

f(g(x))\,g'(x)

or differs from it only by a constant factor. When this pattern appears, choosing $u = g(x)$ simplifies the expression by reducing the composite structure to a single variable. A common signal is the presence of expressions such as $(ax+b)^n$, $\sqrt{ax+b}$, $\ln(ax+b)$, or $e^{ax+b}$. In these cases, the inner linear function $ax+b$ is often a natural candidate for substitution. Similarly, in rational expressions of the form:

\frac{g'(x)}{g(x)}

the derivative of the denominator suggests the substitution $u = g(x)$.

In practice, the key idea is to look for an inner expression whose derivative also appears, exactly or up to a multiplicative constant, elsewhere in the integrand. When such a relationship is present, substitution typically transforms the integral into a simpler form.

Substitution patterns

\[ \int f(g(x))\, g'(x)\, dx \]\[ u = g(x) \]
\[ \int (ax+b)^n\, dx \]\[ u = ax+b \]
\[ \int e^{ax+b}\]\[ u = ax+b \]
\[\int \ln(ax+b)\, dx \]\[ u = ax+b \]
\[ \int \dfrac{g'(x)}{g(x)}\, dx \]\[ u = g(x) \]

Example 1

Consider the following integral:

\int (2x+1)^3 \,dx

Let $u = 2x + 1$, which simplifies the exponentiation. Differentiating both sides with respect to $x$ we have:

du = 2\,dx

Solving for $dx$ we obtain:

dx = \frac{du}{2}

Expressing the integral entirely in terms of $u$:

\int u^3 \cdot \frac{du}{2} = \frac{1}{2} \int u^3 \,du

We now proceed to solve the integral in $u$, which has been reduced to a power integral of the form $u^a$. We obtain:

\frac{1}{2} \cdot \frac{u^4}{4} + c = \frac{1}{8} u^4 + c

Substituting back $u = 2x+1$, we get:

\frac{1}{8} (2x+1)^4 + c

Example 2

Evaluate the following integral:

\int \frac{1}{3x-5} \,dx

Let $u = 3x - 5$, which simplifies the denominator. Differentiating both sides with respect to $x$, we have:

du = 3\,dx

Solving for $dx$, we obtain:

dx = \frac{du}{3}

Expressing the integral entirely in terms of $u$:

\int \frac{1}{u} \cdot \frac{du}{3} = \frac{1}{3} \int \frac{du}{u}

We now proceed to solve the integral in $u$, which has been reduced to a logarithmic integral of the form $1/u$. We obtain:

\frac{1}{3} \ln |u| + c

Substituting back $u = 3x - 5$, we get:

\frac{1}{3} \ln |3x - 5| + c

Integration by substitution is an effective technique, but selecting the right substitution requires practice and the ability to recognize the structure of the integrand.

Example 3

Evaluate the following integral:

\int x \sin(x^2)\,dx

The substitution is less immediate here, since the integrand does not match the standard pattern as directly as in the previous examples. Let $u = x^2$, which simplifies the argument of the sine function. Differentiating both sides with respect to $x$, we get:

du = 2x\,dx

Solving for $dx$ we obtain:

dx = \frac{du}{2x}

Rewriting everything in terms of $u$, and since $du = 2x\,dx$, we have:

\int x \sin(u) \cdot \frac{du}{2x} = \frac{1}{2} \int \sin(u)\,du

We now proceed to solve the integral in $u$:

\int \sin u\,du = -\cos u

Thus:

\frac{1}{2} (-\cos u) + c = -\frac{1}{2} \cos u + c

Substituting back $u = x^2$, we obtain:

-\frac{1}{2} \cos(x^2) + c

Example 4

Evaluate the following integral:

\int \cos x \sqrt{\sin x}\,dx

Let $u = \sin x$, which simplifies the square root term. Differentiating both sides with respect to $x$, we get:

du = \cos x\,dx

Since $du = \cos x\,dx$, we can substitute directly into the integral.

Substituting $u = \sin x$ and $du = \cos x\,dx$ into the integral, we obtain:

\int \sqrt{u}\,du = \int u^{1/2}\,du

We now compute the integral:

\int u^{1/2}\,du = \frac{u^{3/2}}{\frac{3}{2}} = \frac{2}{3}u^{3/2} + c

Substituting back $u = \sin x$, we obtain:

\frac{2}{3} (\sin x)^{3/2} + c

Trigonometric substitutions

Trigonometric substitution applies when an integral involves polynomial, rational, or algebraic expressions that can be simplified using the fundamental trigonometric identity:

\sin^2 x + \cos^2 x = 1

which can be expressed in the following forms:

\begin{align*} \cos^2 x &= 1 - \sin^2 x \\[0.5em] \sec^2 x &= 1 + \tan^2 x \\[0.5em] \tan^2 x &= \sec^2 x - 1 \end{align*}

To simplify an integral, choose an appropriate substitution based on the expression present in the integrand:

  • If the integrand contains $1 - x^2$, use $x = \sin u$.
  • If the integrand contains $1 + x^2$, use $x = \tan u$.
  • If the integrand contains $x^2 - 1$, use $x = \sec u$.

A complete discussion of trigonometric substitution, including the geometric rationale and fully worked examples, is presented in the dedicated section Trigonometric Substitution for Integrals.

Example 5

Evaluate the following integral:

\int \frac{1}{\sqrt{9-x^2}}\,dx

For expressions of the form $a^2 - x^2$, a natural substitution is:

x = 3\sin u

Differentiating both sides:

dx = 3\cos u\,du

Substituting $x = 3\sin u$ in the denominator:

\sqrt{9-x^2} = \sqrt{9-9\sin^2 u} = \sqrt{9(1-\sin^2 u)}

Since $\sin^2 u + \cos^2 u = 1$, we obtain:

\sqrt{9(1-\sin^2 u)} = \sqrt{9\cos^2 u} = 3\cos u

Thus, the integral transforms into:

\int \frac{3\cos u\,du}{3\cos u} = \int du = u+c

This step assumes $\cos u \geq 0$, which holds since the substitution $x = 3\sin u$ implies $u \in [-\pi/2,\,\pi/2]$.

From the substitution $x = 3\sin u$, solving for $u$ via the arcsine function gives:

u = \arcsin\left(\frac{x}{3}\right)

Thus:

\arcsin\left(\frac{x}{3}\right) + c

Substitution rule for definite integrals

When applying substitution to evaluate definite integrals, the limits of integration must be adjusted to reflect the new variable. If the limits are not changed, the result will be incorrect. Given the substitution $u = g(x)$, we have:

\int_{a}^{b} f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du

Example 6

Evaluate the following definite integral:

\int_{0}^{1} x\cos(x^2)\,dx

Using the substitution $u = x^2$, we get:

du = 2x\,dx \qquad dx = \frac{du}{2x}

The limits of integration must be updated. When $x = 0$, then $u = 0$; when $x = 1$, then $u = 1$. In this case the transformed limits coincide with the original ones, though this is not generally the case. We obtain:

\int_{0}^{1} x\cos(x^2)\,dx = \int_{0}^{1} \cos u \cdot \frac{du}{2} = \frac{1}{2}\int_{0}^{1} \cos u\,du

Evaluating the integral we obtain:

\frac{1}{2}\sin u\Big|_{0}^{1} = \frac{1}{2}(\sin 1 - \sin 0) = \frac{\sin 1}{2}

Flowchart

  • Integral to solve
    • IF the integrand matches a known form
      • integrate directly apply the corresponding standard formula (power, exponential, trigonometric…)
    • ELSE IF the integrand contains $a^2 - x^2$ or $a^2 + x^2$ or $x^2 - a^2$
      • choose the trigonometric substitution
\begin{align*} a^2 - x^2 &\rightarrow x = a\sin u \\[3pt] a^2 + x^2 &\rightarrow x = a\tan u \\[3pt] x^2 - a^2 &\rightarrow x = a\sec u \end{align*}
- _compute $dx$ from the substitution_
  differentiate both sides: e.g. $x = a\sin u \rightarrow dx = a\cos u\,du$
- _apply the Pythagorean identity_
  the radical collapses to a single trigonometric function
- _integrate in $u$_
  the result is now a standard trigonometric integral
- **IF** the integral is definite
  - _update the limits of integration_
    replace $a$ and $b$ with the corresponding values of $u$ before evaluating
  - _evaluate and stop_
    the result is a number, no back-substitution needed
- `ELSE`
  - _back-substitute using the inverse function_
    express the result in terms of the original variable $x$
  • ELSE IF the integrand has the form $f(g(x)) \cdot g'(x)$
    • set $u = g(x)$ choose the inner function whose derivative appears in the integrand
    • compute $du = g'(x)\,dx$ if needed, solve for $dx$ and adjust for any constant factor
    • rewrite the integral entirely in $u$ every occurrence of $x$ and $dx$ must be replaced
    • integrate in $u$ apply the appropriate standard formula
    • IF the integral is definite
      • update the limits of integration replace $a$ and $b$ with $g(a)$ and $g(b)$ before evaluating
      • evaluate and stop the result is a number, no back-substitution needed
    • ELSE
      • substitute back $u = g(x)$ express the antiderivative in terms of the original variable $x$
  • ELSE
    • simplify the integrand first expand, factor, use trigonometric identities, or split into partial fractions
    • restart