Radical Substitution Method | Indefinite Integrals

1. Method Principles

Definition 1.1: Radical Substitution Method

When the integrand contains a radical $\sqrt[n]{f(x)}$ , use appropriate substitution to eliminate the radical and simplify the integral.

Core idea: Let $t = \sqrt[n]{f(x)}$ or use trigonometric substitution to convert radicals into rational or trigonometric expressions.

Substitution Strategy

Goal: Through substitution, make the integral in the new variable simpler

Key: Choose an appropriate substitution function so that the expression for dx simplifies as well

Note: After substitution, check the domain and remember to back-substitute to the original variable

2. Classification of Radical Substitution

Theorem 2.1: Linear Radical Substitution

For integrals containing $\sqrt[n]{ax+b}$ :

Substitution method: Let $t = \sqrt[n]{ax+b}$

Then $ax+b = t^n$ , $x = \frac{t^n-b}{a}$ , $dx = \frac{n}{a}t^{n-1}dt$

Theorem 2.2: Trigonometric Substitution for Quadratic Radicals

For integrals containing quadratic radicals, use trigonometric substitution:

Type 1: $\sqrt{a^2-x^2}$

Let $x = a\sin\theta$ , then $\sqrt{a^2-x^2} = a\cos\theta$

Applicable range: $|x| \leq a$

Type 2: $\sqrt{x^2+a^2}$

Let $x = a\tan\theta$ , then $\sqrt{x^2+a^2} = a\sec\theta$

Applicable range: All real numbers $x$

Type 3: $\sqrt{x^2-a^2}$

Let $x = a\sec\theta$ , then $\sqrt{x^2-a^2} = a\tan\theta$

Applicable range: $|x| \geq a$

3. Detailed Example Solutions

Example 3.1: Basic Application of Linear Radicals

Problem: Find $\int \sqrt{\frac{x}{x-1}}\,dx$

Solution:

Step 1: Rewrite the radical

\sqrt{\frac{x}{x-1}} = \sqrt{\frac{x-1+1}{x-1}} = \sqrt{1 + \frac{1}{x-1}}

Step 2: Substitution

Let $t = \sqrt{x-1}$ , then $x = t^2 + 1$ , $dx = 2t\,dt$

\begin{aligned} \sqrt{\frac{x}{x-1}} &= \sqrt{\frac{t^2+1}{t^2}} = \frac{\sqrt{t^2+1}}{|t|} \end{aligned}

Step 3: Integration

\begin{aligned} I &= \int \frac{\sqrt{t^2+1}}{t} \cdot 2t\,dt = 2\int\sqrt{t^2+1}\,dt \end{aligned}

This is a standard $\sqrt{x^2+a^2}$ type integral, using the formula:

\int\sqrt{t^2+1}\,dt = \frac{t}{2}\sqrt{t^2+1} + \frac{1}{2}\ln|t+\sqrt{t^2+1}| + C

Step 4: Back substitution

I = t\sqrt{t^2+1} + \ln|t+\sqrt{t^2+1}| + C

= \sqrt{x-1}\cdot\sqrt{x} + \ln|\sqrt{x-1}+\sqrt{x}| + C

Answer

I = \sqrt{x(x-1)} + \ln\left|\sqrt{x-1}+\sqrt{x}\right| + C

Example 3.2: Trigonometric Substitution: √(a²-x²) Type

Problem: Find $\int \frac{1}{x^2\sqrt{4-x^2}}\,dx$

Solution:

Step 1: Trigonometric substitution

Let $x = 2\sin\theta$ , then:

\begin{aligned} \sqrt{4-x^2} &= \sqrt{4-4\sin^2\theta} = 2\cos\theta \\ dx &= 2\cos\theta\,d\theta \end{aligned}

Step 2: Substitute and integrate

\begin{aligned} I &= \int \frac{1}{4\sin^2\theta \cdot 2\cos\theta} \cdot 2\cos\theta\,d\theta \\ &= \int \frac{1}{4\sin^2\theta}\,d\theta = \frac{1}{4}\int \csc^2\theta\,d\theta \\ &= -\frac{1}{4}\cot\theta + C \end{aligned}

Step 3: Back substitution

From $x = 2\sin\theta$ , we get $\sin\theta = \frac{x}{2}$

Therefore $\cos\theta = \sqrt{1-\sin^2\theta} = \frac{\sqrt{4-x^2}}{2}$

\cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\sqrt{4-x^2}}{x}

I = -\frac{1}{4}\cdot\frac{\sqrt{4-x^2}}{x} + C = -\frac{\sqrt{4-x^2}}{4x} + C

Answer

I = -\frac{\sqrt{4-x^2}}{4x} + C

Example 3.3: Trigonometric Substitution: √(x²+a²) Type

Problem: Find $\int \frac{1}{(1+x^2)\sqrt{x}}\,dx$

Solution:

Step 1: First substitution

First handle $\sqrt{x}$ , let $u = \sqrt{x}$ , then $x = u^2$ , $dx = 2u\,du$

I = \int \frac{1}{(1+u^4)u} \cdot 2u\,du = 2\int \frac{1}{1+u^4}\,du

Step 2: Factor the denominator

Observe that:

1 + u^4 = (1+\sqrt{2}u+u^2)(1-\sqrt{2}u+u^2)

Use partial fraction decomposition (detailed steps omitted):

\frac{1}{1+u^4} = \frac{1}{2\sqrt{2}}\left(\frac{1}{u^2-\sqrt{2}u+1} - \frac{1}{u^2+\sqrt{2}u+1}\right)

Step 3: Integrate separately

Complete the square for both terms, then integrate using arctangent and logarithm formulas

Step 4: Back-substitute

Substitute $u = \sqrt{x}$ back into the final result

Note

This problem involves complex calculations. In practice, you may need to consult tables or use a computer algebra system.

Example 3.4: Combining Radicals and Trigonometric Functions

Problem: Find $\int \frac{1}{x\sqrt{2x^2+2x-1}}\,dx$

Solution:

Step 1: Complete the square

2x^2+2x-1 = 2\left(x^2+x-\frac{1}{2}\right) = 2\left[\left(x+\frac{1}{2}\right)^2 - \frac{3}{4}\right]

Step 2: Substitution

Let $u = x + \frac{1}{2}$ , then $x = u - \frac{1}{2}$ , $dx = du$

I = \int \frac{1}{(u-\frac{1}{2})\sqrt{2u^2-\frac{3}{2}}}\,du

Further simplification:

= \int \frac{1}{(u-\frac{1}{2})\sqrt{2}\sqrt{u^2-\frac{3}{4}}}\,du

Step 3: Trigonometric or hyperbolic substitution

This requires more advanced techniques, typically involving hyperbolic functions or consulting integral tables

Difficulty Note

This type of problem involves complex combinations of radicals and rational functions, typically requiring multiple substitutions or table consultation.

Example 3.5: Special Radical Techniques

Problem: Find $\int x\sqrt{2x-x^2}\,dx$

Solution:

Step 1: Complete the square

2x - x^2 = -(x^2-2x) = -(x^2-2x+1-1) = 1-(x-1)^2

So $\sqrt{2x-x^2} = \sqrt{1-(x-1)^2}$

Step 2: Substitution

Let $x-1 = \sin\theta$ , then $x = 1+\sin\theta$ , $dx = \cos\theta\,d\theta$

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

Step 3: Substitute and integrate

\begin{aligned} I &= \int (1+\sin\theta)\cos\theta \cdot \cos\theta\,d\theta \\ &= \int (1+\sin\theta)\cos^2\theta\,d\theta \\ &= \int \cos^2\theta\,d\theta + \int \sin\theta\cos^2\theta\,d\theta \end{aligned}

Step 4: Calculate separately

First term:

\int \cos^2\theta\,d\theta = \int \frac{1+\cos 2\theta}{2}\,d\theta = \frac{\theta}{2} + \frac{\sin 2\theta}{4} + C_1

Second term:

\int \sin\theta\cos^2\theta\,d\theta = -\frac{\cos^3\theta}{3} + C_2

Step 5: Combine and back substitute

I = \frac{\theta}{2} + \frac{\sin 2\theta}{4} - \frac{\cos^3\theta}{3} + C

From $\sin\theta = x-1$ , we get $\theta = \arcsin(x-1)$ , $\cos\theta = \sqrt{2x-x^2}$

Answer

I = \frac{1}{2}\arcsin(x-1) + \frac{(x-1)\sqrt{2x-x^2}}{2} - \frac{(2x-x^2)^{3/2}}{3} + C

4. Trigonometric Substitution Quick Reference

Radical Type	Substitution	Radical Simplifies To	dx Expression
$\sqrt{a^2-x^2}$	$x = a\sin\theta$	$a\cos\theta$	$a\cos\theta\,d\theta$
$\sqrt{a^2+x^2}$	$x = a\tan\theta$	$a\sec\theta$	$a\sec^2\theta\,d\theta$
$\sqrt{x^2-a^2}$	$x = a\sec\theta$	$a\tan\theta$	$a\sec\theta\tan\theta\,d\theta$

5. Problem-Solving Techniques and Precautions

Preparation Before Substitution

• Observe the radical form and determine which type it belongs to

• If the radical contains a linear term, complete the square first to get standard form

• Check whether the integrand can be simplified beforehand

• Note domain restrictions

Trigonometric Substitution Techniques

• Draw a right triangle to help remember trigonometric relationships

• Pay attention to the angle range (usually take principal values)

• Use triangle relationships when back-substituting to avoid tedious calculations

• Final result should not contain trigonometric functions

Common Mistakes

• Forgetting to calculate the expression for dx

• Making errors in trigonometric relationships during back-substitution

• Ignoring absolute value signs

• Incorrect domain determination

Verifying Your Answer

• Differentiate the result to verify it equals the integrand

• Check that the constant of integration is included

• Confirm the answer's domain is correct

• Test with specific values