Radicals and the Substitution Pattern
Square roots of polynomial expressions often signal a substitution. The reason is structural: when the radicand is \(g(x)\) and the integrand also contains a multiple of \(g'(x)\), the Chain Rule in reverse applies. The pattern is \(\displaystyle\int_a^b g'(x)\sqrt{g(x)}\,dx\), which becomes \(\displaystyle\int_{g(a)}^{g(b)}\sqrt{u}\,du\) after the substitution \(u=g(x)\). In definite integrals, the limits are converted at the same time, so we never need to return to the variable \(x\).
Worked Example 1: x Times a Radical
Evaluate \(\displaystyle\int_0^2 x\sqrt{x^2+5}\,dx\).
The radicand is \(x^2+5\), whose derivative is \(2x\). The integrand has \(x\), which is half of that derivative. Let \(u=x^2+5\), so \(du=2x\,dx\) and \(x\,dx=\tfrac12\,du\). The limits become \(u(0)=5\) and \(u(2)=9\). Then
Check: \((27-5\sqrt5)/3\approx5.27322\), and numerical integration of the original integral gives 5.27322.
Worked Example 2: A Quadratic Derivative Factor
Evaluate \(\displaystyle\int_0^1(3x^2+1)\sqrt{x^3+x}\,dx\).
The radicand is \(x^3+x\) with derivative \(3x^2+1\), which appears exactly. Let \(u=x^3+x\), so \(du=(3x^2+1)\,dx\). The limits are \(u(0)=0\) and \(u(1)=2\). Then
Check: \(4\sqrt2/3\approx1.88562\), matching numerical integration.
Worked Example 3: A Radical in the Denominator
Evaluate \(\displaystyle\int_1^2\frac{2x+3}{\sqrt{x^2+3x}}\,dx\).
With \(u=x^2+3x\), \(du=(2x+3)\,dx\). The limits are \(u(1)=4\) and \(u(2)=10\). So
Check: \(2(\sqrt{10}-2)\approx2.32456\), and the numerical integral agrees.
Worked Example 4: Square Root of a Sum of Squares
Evaluate \(\displaystyle\int_0^4\frac{x}{\sqrt{x^2+9}}\,dx\).
Let \(u=x^2+9\); \(x\,dx=\tfrac12du\); the limits are \(9\) and \(25\). Then
The numerical integral of the original function over \([0,4]\) is also \(2\), and the nice integer answer comes from the perfect squares \(9\) and \(25\).
Worked Example 5: A Cubic Radicand
Evaluate \(\displaystyle\int_0^1x^2\sqrt{x^3+1}\,dx\).
Take \(u=x^3+1\), \(x^2dx=\tfrac13du\); the limits are \(1\) and \(2\). Then
Check: the value is 0.406317, matching numerical integration.
Summary of the Method
| Step | Action | Example 1 |
|---|---|---|
| 1 | Identify the radicand g(x) and look for g'(x) or a multiple in the integrand | g = x² + 5, g' = 2x |
| 2 | Set u = g(x) and write du | du = 2x dx |
| 3 | Adjust the constant factor | x dx = ½ du |
| 4 | Convert both limits | x = 0 → u = 5; x = 2 → u = 9 |
| 5 | Integrate in u and evaluate | (27 − 5√5)/3 |
Common Mistakes
Forgetting to change the limits. If the limits \(0\) and \(2\) are kept after substituting, the integral \(\tfrac12\int_0^2\sqrt u\,du=\tfrac13\,2^{3/2}\approx0.9428\) is obtained, but the correct value of Example 1 is about 5.2732.
Missing constants. In Example 1, \(x\,dx\) is \(\tfrac12\,du\), not \(du\). Omitting the factor \(\tfrac12\) doubles the answer. In Example 5, the constant is \(\tfrac13\).
Returning to \(x\) after converting the limits. Once the limits are in terms of \(u\), the answer is a number. Substituting \(x^2+5\) back for \(u\) and then using the \(u\) limits mixes the two variables.
Why the Limits Can Be Converted
Let \(F\) be an antiderivative of \(f\). By the Chain Rule, \(\dfrac{d}{dx}F(g(x))=f(g(x))\,g'(x)\), so \(F\circ g\) is an antiderivative of \(f(g(x))g'(x)\). The Fundamental Theorem then gives \(\int_a^b f(g(x))g'(x)\,dx=F(g(b))-F(g(a))\). But the right side is exactly \(\int_{g(a)}^{g(b)}f(u)\,du\). This is the entire justification for converting the limits: the values of \(u\) at the two ends of the interval are \(g(a)\) and \(g(b)\), and the evaluation takes place in the new variable. No back-substitution is required.
Worked Example 6: A Linear Factor Hiding the Derivative
Evaluate \(\displaystyle\int_0^2(x+1)\sqrt{x^2+2x+4}\,dx\).
The radicand \(x^2+2x+4\) has derivative \(2x+2=2(x+1)\), so the factor \(x+1\) is half of the derivative. Let \(u=x^2+2x+4\), so \((x+1)\,dx=\tfrac12du\). The limits are \(u(0)=4\) and \(u(2)=12\). Then
Check: the exact value is \(\approx11.1897\), which agrees with numerical integration.
Estimating Before You Compute
A rough estimate guards against arithmetic slips. In Example 1, the integrand \(x\sqrt{x^2+5}\) equals \(0\) at \(x=0\) and \(2\cdot3=6\) at \(x=2\). It increases throughout, so the area over an interval of length \(2\) must be between \(0\) and \(12\), and since the curve is slightly bowed, an average height near \(2.6\) is plausible. The exact value \(\tfrac13(27-5\sqrt5)\approx5.273\) corresponds to an average height of about 2.64, which is consistent. If a computation had returned a number such as \(10.5\) or \(-1\), the estimate would have flagged it immediately.
When the Pattern Is Not Present
The method works because the derivative of the radicand appears in the integrand. Compare \(\int_0^2x\sqrt{x^2+5}\,dx\) with \(\int_0^2\sqrt{x^2+5}\,dx\). In the second integral there is no factor \(x\), so \(u=x^2+5\) does not work: the leftover \(dx\) would be \(\dfrac{du}{2x}=\dfrac{du}{2\sqrt{u-5}}\), which does not simplify. That integral requires a trigonometric or hyperbolic substitution, a different technique. Likewise, \(\int x^2\sqrt{x^2+5}\,dx\) has an extra factor of \(x\) that cannot be absorbed. The first step in any such problem is to ask whether the derivative of the radicand, up to a constant factor, appears in the integrand.
Worked Example 7: A Decreasing Radicand
Evaluate \(\displaystyle\int_0^1x\sqrt{1-x^2}\,dx\).
Let \(u=1-x^2\), so \(du=-2x\,dx\) and \(x\,dx=-\tfrac12du\). The limits are \(u(0)=1\) and \(u(1)=0\), so the new integral runs from \(1\) down to \(0\):
The minus sign from \(du\) is absorbed by reversing the limits. This is a frequent source of sign errors: when \(u=g(x)\) is decreasing, the lower limit in \(u\) is larger than the upper limit. The numerical integral of the original function is \(0.3333\), confirming the answer.
Interpreting the Result
The integral in Example 4 equals the area under \(y=x/\sqrt{x^2+9}\) from \(x=0\) to \(x=4\). Since the integral is \(2\) and the interval has length \(4\), the average value of the function on the interval is \(\tfrac24=0.5\). The function increases from \(0\) at \(x=0\) to \(\tfrac45=0.8\) at \(x=4\), and an average of \(0.5\) lies between these two values, as it must. Interpreting an answer as an average is a quick sanity check that works for any definite integral of a continuous function: the average must lie between the minimum and maximum of the integrand.
The same check applies to Example 7. The integrand \(x\sqrt{1-x^2}\) is \(0\) at both ends of \([0,1]\) and reaches its maximum value \(\tfrac12\) at \(x=\tfrac1{\sqrt2}\). An integral of \(\tfrac13\) over an interval of length \(1\) means an average height of \(\tfrac13\), which is below the maximum of \(\tfrac12\), as expected.
Where This Fits in the Course
This tutorial belongs to the module “Substitution and Reversing the Chain Rule”. The tutorial just before it is “Rational Definite Integrals By Substitution,” and the one that follows is “Evaluating Without Solving For The Original Variable.”
Practice Problems with Answers
Try each problem before reading the answer. Each answer below was verified by differentiation or by evaluating a difference quotient or integral numerically.
- Evaluate \(\int_0^3 x\sqrt{x^2+16}\,dx\).
Answer. \(\tfrac13(25^{3/2}-16^{3/2})=\tfrac13(125-64)=\tfrac{61}{3}\). - Evaluate \(\int_1^2 (2x)\sqrt{x^2+1}\,dx\).
Answer. \(\tfrac23(5^{3/2}-2^{3/2})\approx5.5679\). - Evaluate \(\int_0^1\dfrac{x}{\sqrt{x^2+1}}dx\).
Answer. \(\sqrt2-1\). - Evaluate \(\int_0^2 (x^2)\sqrt{x^3+1}\,dx\).
Answer. \(\tfrac29(27-1)=\tfrac{52}{9}\).
Check Your Understanding
Work these without looking back.
- Why do the limits change when we substitute in a definite integral?
- Evaluate \(\int_0^1 x\sqrt{1+x^2}\,dx\).
- Evaluate \(\int_1^3\dfrac{2x}{\sqrt{x^2+7}}\,dx\).
- Evaluate \(\int_0^2(x+1)\sqrt{x^2+2x+4}\,dx\).
- Explain how the substitution pattern is the Chain Rule in reverse.
