Special Substitutions I

In our previous tutorial, we explored the mechanics of substitution and the algebra of differentials.

Now we turn to a central skill in indefinite integration: knowing which substitution to choose when faced with specific algebraic and radical forms.

Many integrals contain expressions involving square roots that cannot be simplified by polynomial algebra. However, by replacing \(x\) with an appropriate trigonometric function, the Pythagorean identities dissolve the square roots completely, turning difficult integrands into standard table forms.

Standard Trigonometric Substitutions

Every trigonometric substitution relies on a fundamental identity to convert a two-term sum or difference under a square root into a single squared term.

Here is an exhaustive reference of the primary patterns:

Algebraic Expression	Recommended Substitution	Differential \(dx\)	Key Identity Used	Simplified Radical
\(\sqrt{a^2 - x^2}\)	\(x = a\sin\theta\)	\(dx = a\cos\theta\,d\theta\)	\(1 - \sin^2\theta = \cos^2\theta\)	\(a\cos\theta\)
\(\sqrt{a^2 + x^2}\)	\(x = a\tan\theta\)	\(dx = a\sec^2\theta\,d\theta\)	\(1 + \tan^2\theta = \sec^2\theta\)	\(a\sec\theta\)
\(\sqrt{x^2 - a^2}\)	\(x = a\sec\theta\)	\(dx = a\sec\theta\tan\theta\,d\theta\)	\(\sec^2\theta - 1 = \tan^2\theta\)	\(a\tan\theta\)
\(\sqrt{\frac{a - x}{a + x}}\) or \(\sqrt{\frac{a + x}{a - x}}\)	\(x = a\cos 2\theta\)	\(dx = -2a\sin 2\theta\,d\theta\)	\(\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \tan^2\theta\)	\(\tan\theta\)
\(\sqrt{(x - a)(b - x)}\)	\(x = a\cos^2\theta + b\sin^2\theta\)	\(dx = (b - a)\sin 2\theta\,d\theta\)	Factors become squares	\((b - a)\sin\theta\cos\theta\)
\(\sqrt{(x - a)(x - b)}\)	\(x = a\sec^2\theta - b\tan^2\theta\)	\(dx = 2(a - b)\sec^2\theta\tan\theta\,d\theta\)	\(\sec^2\theta - \tan^2\theta = 1\)	\((a - b)\sec\theta\tan\theta\)

Example 1

Evaluate:

\[I = \int \frac{dx}{x^2\sqrt{a^2 - x^2}}\]

Solution:

The presence of \(\sqrt{a^2 - x^2}\) suggests the substitution \(x = a\sin\theta\).

Let:

\[x = a\sin\theta \implies dx = a\cos\theta\,d\theta\]

Under this substitution:

\[\sqrt{a^2 - x^2} = \sqrt{a^2(1 - \sin^2\theta)} = a\cos\theta\]

Substitute into the integral:

\[I = \int \frac{a\cos\theta\,d\theta}{(a^2\sin^2\theta)(a\cos\theta)} = \frac{1}{a^2}\int \frac{d\theta}{\sin^2\theta} = \frac{1}{a^2}\int \csc^2\theta\,d\theta\]

Integrating \(\csc^2\theta\):

\[I = -\frac{1}{a^2}\cot\theta + C\]

To express the result in terms of \(x\), draw a right-angled triangle with angle \(\theta\), opposite side \(x\), and hypotenuse \(a\). The adjacent side is \(\sqrt{a^2 - x^2}\).

Therefore:

\[\cot\theta = \frac{\sqrt{a^2 - x^2}}{x}\]

Substituting this back gives:

\[I = -\frac{\sqrt{a^2 - x^2}}{a^2 x} + C\]

Example 2

Evaluate:

\[I = \int \frac{dx}{(a^2 + x^2)^{3/2}}\]

Solution:

The term \(a^2 + x^2\) matches the Pythagorean identity \(1 + \tan^2\theta = \sec^2\theta\).

Let:

\[x = a\tan\theta \implies dx = a\sec^2\theta\,d\theta\]

Then:

\[a^2 + x^2 = a^2(1 + \tan^2\theta) = a^2\sec^2\theta\]

\[(a^2 + x^2)^{3/2} = (a^2\sec^2\theta)^{3/2} = a^3\sec^3\theta\]

Substitute into the integral:

\[I = \int \frac{a\sec^2\theta\,d\theta}{a^3\sec^3\theta} = \frac{1}{a^2}\int \frac{1}{\sec\theta}\,d\theta = \frac{1}{a^2}\int \cos\theta\,d\theta\]

\[I = \frac{1}{a^2}\sin\theta + C\]

From \(x = a\tan\theta\), we have \(\tan\theta = \frac{x}{a}\). The opposite side is \(x\), adjacent is \(a\), and the hypotenuse is \(\sqrt{a^2 + x^2}\).

Hence:

\[\sin\theta = \frac{x}{\sqrt{a^2 + x^2}}\]

Therefore:

\[I = \frac{x}{a^2\sqrt{a^2 + x^2}} + C\]

Example 3

Evaluate:

\[I = \int \frac{\sqrt{x^2 - a^2}}{x}\,dx \qquad (x > a > 0)\]

Solution:

The expression \(\sqrt{x^2 - a^2}\) suggests the secant substitution.

Let:

\[x = a\sec\theta \implies dx = a\sec\theta\tan\theta\,d\theta\]

Then:

\[\sqrt{x^2 - a^2} = \sqrt{a^2(\sec^2\theta - 1)} = a\tan\theta\]

Substitute these relations into the integral:

\[I = \int \frac{a\tan\theta}{a\sec\theta}(a\sec\theta\tan\theta\,d\theta) = a\int \tan^2\theta\,d\theta\]

Using the identity \(\tan^2\theta = \sec^2\theta - 1\):

\[I = a\int (\sec^2\theta - 1)\,d\theta = a(\tan\theta - \theta) + C\]

Return to the variable \(x\):

Since \(\sec\theta = \frac{x}{a}\), we have:

\[\tan\theta = \sqrt{\sec^2\theta - 1} = \frac{\sqrt{x^2 - a^2}}{a}\]

\[\theta = \sec^{-1}\left(\frac{x}{a}\right)\]

Substituting these yields:

\[I = a\left( \frac{\sqrt{x^2 - a^2}}{a} - \sec^{-1}\left(\frac{x}{a}\right) \right) + C\]

\[I = \sqrt{x^2 - a^2} - a\sec^{-1}\left(\frac{x}{a}\right) + C\]

Example 4

Evaluate:

\[I = \int \sqrt{\frac{a - x}{a + x}}\,dx \qquad (-a < x < a)\]

Solution:

The ratio \(\frac{a - x}{a + x}\) can be transformed into a single trigonometric ratio using the half-angle identity:

\[\frac{1 - \cos 2\theta}{1 + \cos 2\theta} = \frac{2\sin^2\theta}{2\cos^2\theta} = \tan^2\theta\]

Let:

\[x = a\cos 2\theta \implies dx = -2a\sin 2\theta\,d\theta\]

For \(x \in (-a, a)\), we choose \(\theta \in \left(0, \frac{\pi}{2}\right)\). The radical simplifies to:

\[\sqrt{\frac{a - a\cos 2\theta}{a + a\cos 2\theta}} = \sqrt{\frac{1 - \cos 2\theta}{1 + \cos 2\theta}} = \tan\theta = \frac{\sin\theta}{\cos\theta}\]

Substitute into the integral:

\[I = \int \left(\frac{\sin\theta}{\cos\theta}\right)(-2a\sin 2\theta\,d\theta)\]

Using \(\sin 2\theta = 2\sin\theta\cos\theta\):

\[I = \int \left(\frac{\sin\theta}{\cos\theta}\right)(-4a\sin\theta\cos\theta\,d\theta) = -4a\int \sin^2\theta\,d\theta\]

Apply the identity \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\):

\[I = -2a\int (1 - \cos 2\theta)\,d\theta = -2a\left( \theta - \frac{\sin 2\theta}{2} \right) + C = -2a\theta + a\sin 2\theta + C\]

Return to \(x\):

From \(x = a\cos 2\theta\), we have \(\cos 2\theta = \frac{x}{a}\), which gives:

\[2\theta = \cos^{-1}\left(\frac{x}{a}\right)\]

\[\sin 2\theta = \sqrt{1 - \cos^2 2\theta} = \sqrt{1 - \frac{x^2}{a^2}} = \frac{\sqrt{a^2 - x^2}}{a}\]

Substituting these back:

\[I = -a\cos^{-1}\left(\frac{x}{a}\right) + \sqrt{a^2 - x^2} + C\]

(Note: Since \(\cos^{-1}\left(\frac{x}{a}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{x}{a}\right)\), this antiderivative can equivalently be written as \(a\sin^{-1}\left(\frac{x}{a}\right) + \sqrt{a^2 - x^2} + C_1\).)

Example 5

Evaluate:

\[I = \int \frac{dx}{\sqrt{(x - a)(b - x)}} \qquad (a < x < b)\]

Solution:

When the radical contains the product \((x - a)(b - x)\), a weighted combination of squares decouples both factors simultaneously.

Let:

\[x = a\cos^2\theta + b\sin^2\theta \qquad \left(0 < \theta < \frac{\pi}{2}\right)\]

Notice how each factor simplifies:

\[x - a = a\cos^2\theta + b\sin^2\theta - a = a(\cos^2\theta - 1) + b\sin^2\theta = (b - a)\sin^2\theta\]

\[b - x = b - a\cos^2\theta - b\sin^2\theta = b(1 - \sin^2\theta) - a\cos^2\theta = (b - a)\cos^2\theta\]

Multiplying both expressions:

\[(x - a)(b - x) = (b - a)^2\sin^2\theta\cos^2\theta\]

\[\sqrt{(x - a)(b - x)} = (b - a)\sin\theta\cos\theta\]

Now compute the differential \(dx\):

\[dx = d[a\cos^2\theta + b\sin^2\theta] = [-2a\cos\theta\sin\theta + 2b\sin\theta\cos\theta]\,d\theta = 2(b - a)\sin\theta\cos\theta\,d\theta\]

Substitute both parts into the integral:

\[I = \int \frac{2(b - a)\sin\theta\cos\theta\,d\theta}{(b - a)\sin\theta\cos\theta} = \int 2\,d\theta = 2\theta + C\]

Return to \(x\):

From \(x - a = (b - a)\sin^2\theta\), we have:

\[\sin^2\theta = \frac{x - a}{b - a} \implies \sin\theta = \sqrt{\frac{x - a}{b - a}}\]

\[\theta = \sin^{-1}\sqrt{\frac{x - a}{b - a}}\]

Therefore:

\[I = 2\sin^{-1}\sqrt{\frac{x - a}{b - a}} + C\]

Substitutions Involving Powers of the Variable

In the standard forms above, the terms inside the radical involve linear or quadratic expressions in \(x\).

In many integrals, the expression under the radical involves higher powers or fractional powers, such as:

\[(a^2 - x^4) = a^2 - (x^2)^2, \qquad (a^6 - x^6) = (a^3)^2 - (x^3)^2, \qquad (a^3 - x^3) = (a^{3/2})^2 - (x^{3/2})^2\]

The guiding principle is:

Identify which power of \(x\) is being squared inside the radical, say \(u = x^n\).
Check whether the remaining algebraic factor in the integrand provides the differential \(x^{n-1}\,dx\).
Apply the appropriate trigonometric substitution directly to \(x^n\).

Example 6

Evaluate:

\[I = \int x \sqrt{\frac{a^2 - x^2}{a^2 + x^2}}\,dx \qquad (0 < x < a)\]

Solution:

If the expression were \(\sqrt{\frac{a - x}{a + x}}\), we would set \(x = a\cos 2\theta\).

Here, the terms inside the radical involve \(x^2\) rather than \(x\). Notice that the factor \(x\,dx\) outside is proportional to the differential of \(x^2\).

Let:

\[x^2 = a^2\cos 2\theta \qquad \left(0 < \theta < \frac{\pi}{4}\right)\]

Differentiating both sides:

\[2x\,dx = -2a^2\sin 2\theta\,d\theta \implies x\,dx = -a^2\sin 2\theta\,d\theta\]

Now simplify the radical:

\[\sqrt{\frac{a^2 - x^2}{a^2 + x^2}} = \sqrt{\frac{a^2(1 - \cos 2\theta)}{a^2(1 + \cos 2\theta)}} = \sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}} = \frac{\sin\theta}{\cos\theta}\]

Substitute into the integral:

\[I = \int \left( \frac{\sin\theta}{\cos\theta} \right) (-a^2\sin 2\theta\,d\theta)\]

Using \(\sin 2\theta = 2\sin\theta\cos\theta\):

\[I = \int \left( \frac{\sin\theta}{\cos\theta} \right) (-2a^2\sin\theta\cos\theta\,d\theta) = -2a^2\int \sin^2\theta\,d\theta\]

Apply the half-angle formula \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\):

\[I = -a^2\int (1 - \cos 2\theta)\,d\theta = -a^2\left( \theta - \frac{\sin 2\theta}{2} \right) + C\]

\[I = -a^2\theta + \frac{a^2}{2}\sin 2\theta + C\]

Return to the variable \(x\):

Since \(\cos 2\theta = \frac{x^2}{a^2}\), we have \(2\theta = \cos^{-1}\left(\frac{x^2}{a^2}\right)\), so \(\theta = \frac{1}{2}\cos^{-1}\left(\frac{x^2}{a^2}\right)\).

Also:

\[\sin 2\theta = \sqrt{1 - \cos^2 2\theta} = \sqrt{1 - \left(\frac{x^2}{a^2}\right)^2} = \frac{\sqrt{a^4 - x^4}}{a^2}\]

Substituting these back:

\[I = -\frac{a^2}{2}\cos^{-1}\left(\frac{x^2}{a^2}\right) + \frac{1}{2}\sqrt{a^4 - x^4} + C\]

Example 7

Evaluate:

\[I = \int \frac{x^2}{\sqrt{a^6 - x^6}}\,dx \qquad (|x| < a)\]

Solution:

Rewrite the expression under the radical as a difference of squares:

\[a^6 - x^6 = (a^3)^2 - (x^3)^2\]

The numerator contains \(x^2\,dx\), which is proportional to the derivative of \(x^3\).

Let:

\[x^3 = a^3\sin\theta \qquad \left(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\right)\]

Differentiating gives:

\[3x^2\,dx = a^3\cos\theta\,d\theta \implies x^2\,dx = \frac{a^3}{3}\cos\theta\,d\theta\]

Rewrite the radical:

\[\sqrt{a^6 - x^6} = \sqrt{a^6 - (a^3\sin\theta)^2} = \sqrt{a^6(1 - \sin^2\theta)} = a^3\cos\theta\]

Substitute into the integral:

\[I = \int \frac{\frac{a^3}{3}\cos\theta\,d\theta}{a^3\cos\theta} = \frac{1}{3}\int 1\,d\theta = \frac{1}{3}\theta + C\]

Return to \(x\):

Since \(x^3 = a^3\sin\theta\), we have \(\sin\theta = \frac{x^3}{a^3}\), so \(\theta = \sin^{-1}\left(\frac{x^3}{a^3}\right)\).

Therefore:

\[I = \frac{1}{3}\sin^{-1}\left(\frac{x^3}{a^3}\right) + C\]

Example 8

Evaluate:

\[I = \int \sqrt{\frac{x}{a^3 - x^3}}\,dx \qquad (0 < x < a)\]

Solution:

Rewrite the integrand:

\[\frac{\sqrt{x}}{\sqrt{(a^{3/2})^2 - (x^{3/2})^2}}\]

Notice that the numerator is \(x^{1/2}\). Differentiating \(x^{3/2}\) gives:

\[\frac{d}{dx}[x^{3/2}] = \frac{3}{2}x^{1/2} = \frac{3}{2}\sqrt{x}\]

This reveals that \(x^{3/2}\) is the variable being squared inside the radical.

Let:

\[x^{3/2} = a^{3/2}\sin\theta \qquad \left(0 < \theta < \frac{\pi}{2}\right)\]

Differentiating gives:

\[\frac{3}{2}x^{1/2}\,dx = a^{3/2}\cos\theta\,d\theta \implies \sqrt{x}\,dx = \frac{2}{3}a^{3/2}\cos\theta\,d\theta\]

Simplify the radical:

\[\sqrt{a^3 - x^3} = \sqrt{(a^{3/2})^2 - (a^{3/2}\sin\theta)^2} = a^{3/2}\cos\theta\]

Substitute both parts into the integral:

\[I = \int \frac{\frac{2}{3}a^{3/2}\cos\theta\,d\theta}{a^{3/2}\cos\theta} = \frac{2}{3}\int 1\,d\theta = \frac{2}{3}\theta + C\]

Return to \(x\):

\[\sin\theta = \frac{x^{3/2}}{a^{3/2}} = \left(\frac{x}{a}\right)^{3/2} \implies \theta = \sin^{-1}\left[ \left(\frac{x}{a}\right)^{3/2} \right]\]

Therefore:

\[I = \frac{2}{3}\sin^{-1}\left[ \left(\frac{x}{a}\right)^{3/2} \right] + C\]

Example 9

Evaluate:

\[I = \int \frac{x}{\sqrt{1 + x^4}}\,dx\]

Solution:

The radical contains \(1 + x^4 = 1 + (x^2)^2\). The numerator has \(x\,dx = \frac{1}{2}d(x^2)\).

This matches the form \(\sqrt{1 + u^2}\) with \(u = x^2\).

Let:

\[x^2 = \tan\theta \qquad \left(0 < \theta < \frac{\pi}{2}\right)\]

Differentiating gives:

\[2x\,dx = \sec^2\theta\,d\theta \implies x\,dx = \frac{1}{2}\sec^2\theta\,d\theta\]

Rewrite the radical:

\[\sqrt{1 + x^4} = \sqrt{1 + \tan^2\theta} = \sec\theta\]

Substitute into the integral:

\[I = \int \frac{\frac{1}{2}\sec^2\theta\,d\theta}{\sec\theta} = \frac{1}{2}\int \sec\theta\,d\theta\]

From our standard table of integrals:

\[I = \frac{1}{2}\ln|\sec\theta + \tan\theta| + C\]

Return to \(x\):

Since \(\tan\theta = x^2\), we have \(\sec\theta = \sqrt{1 + \tan^2\theta} = \sqrt{1 + x^4}\).

Therefore:

\[I = \frac{1}{2}\ln\left( x^2 + \sqrt{1 + x^4} \right) + C\]

Example 10

Evaluate:

\[I = \int \sqrt{\frac{x}{a - x}}\,dx \qquad (0 < x < a)\]

Solution:

Here the numerator has \(x\) and the denominator has \(a - x\).

If we choose a substitution where \(x\) is proportional to \(\sin^2\theta\), then \(a - x\) will naturally become proportional to \(\cos^2\theta\), and their quotient will be \(\tan^2\theta\).

Let:

\[x = a\sin^2\theta \qquad \left(0 < \theta < \frac{\pi}{2}\right)\]

Differentiating gives:

\[dx = 2a\sin\theta\cos\theta\,d\theta\]

Now evaluate the radical:

\[\sqrt{\frac{x}{a - x}} = \sqrt{\frac{a\sin^2\theta}{a - a\sin^2\theta}} = \sqrt{\frac{a\sin^2\theta}{a\cos^2\theta}} = \frac{\sin\theta}{\cos\theta}\]

Substitute into the integral:

\[I = \int \left( \frac{\sin\theta}{\cos\theta} \right) (2a\sin\theta\cos\theta\,d\theta)\]

The \(\cos\theta\) cancels:

\[I = 2a\int \sin^2\theta\,d\theta\]

Using \(\sin^2\theta = \frac{1 - \cos 2\theta}{2}\):

\[I = a\int (1 - \cos 2\theta)\,d\theta = a\left( \theta - \frac{\sin 2\theta}{2} \right) + C = a(\theta - \sin\theta\cos\theta) + C\]

Return to \(x\):

From \(x = a\sin^2\theta\):

\[\sin\theta = \sqrt{\frac{x}{a}} \implies \theta = \sin^{-1}\sqrt{\frac{x}{a}}\]

\[\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{x}{a}} = \sqrt{\frac{a - x}{a}}\]

Therefore:

\[\sin\theta\cos\theta = \sqrt{\frac{x}{a}} \cdot \sqrt{\frac{a - x}{a}} = \frac{\sqrt{x(a - x)}}{a}\]

Substitute these into our antiderivative:

\[I = a\left( \sin^{-1}\sqrt{\frac{x}{a}} - \frac{\sqrt{x(a - x)}}{a} \right) + C\]

\[I = a\sin^{-1}\sqrt{\frac{x}{a}} - \sqrt{x(a - x)} + C\]