Why Bounds Matter
A completed square can turn a complicated quadratic expression into a standard form, such as \(R^2-u^2\) or \(u^2+c^2\). The corresponding antiderivatives may involve inverse trigonometric functions or logarithms. For a definite integral, finding an antiderivative is only part of the work: the radicand must be valid throughout the interval, and a substitution changes the bounds as well as the integrand.
This tutorial focuses on those points. The main forms will be a square root or reciprocal square root of a completed-square quadratic. The formulas used below are applied on intervals where their expressions are defined.
Check the Domain Before Evaluating
For an integrand containing \(\sqrt{q(x)}\), the radicand must satisfy \(q(x)\geq 0\) at every point in the integration interval. If the integrand contains \(1/\sqrt{q(x)}\), it must also satisfy \(q(x)>0\) there for the ordinary continuous-integrand version of the Fundamental Theorem of Calculus to apply. If \(q(x)=0\) at an endpoint, the reciprocal square root is unbounded there and requires separate treatment as an improper integral.
For example, the radicand \(9-(2x-1)^2\) is nonnegative when \(-3\leq 2x-1\leq 3\), or equivalently \(-1\leq x\leq 2\). A proposed interval outside this range cannot be used for a real-valued square root. Checking the whole interval matters: checking only its endpoints is not generally enough unless the shape of the quadratic establishes the sign between them.
Transform Both Bounds
Suppose a substitution is \(u=\alpha(x-h)\), where \(\alpha\neq 0\). Then \(du=\alpha\,dx\), so \(dx=du/\alpha\). An integral from \(x=a\) to \(x=b\) becomes an integral from \(u=\alpha(a-h)\) to \(u=\alpha(b-h)\). These new bounds follow by substituting the original endpoints into the change of variable.
Confirm that it has the required sign throughout the original interval.
Set \(u=\alpha(x-h)\) to make the shifted square the variable’s square.
Compute both new endpoint values and include the factor in \(dx=du/\alpha\).
Use an antiderivative valid on the interval, substitute the bounds, and verify the result.
Why the Substitution Rule Works
The bound transformation is a direct consequence of the chain rule and the Fundamental Theorem of Calculus. Suppose \(G'(u)=g(u)\) on the transformed interval and \(u=\alpha(x-h)\). Define \(F(x)=G(\alpha(x-h))/\alpha\). Differentiating gives \(F'(x)=g(\alpha(x-h))\). Thus \(F\) is an antiderivative of the substituted integrand in \(x\).
This is the same result obtained by writing \(dx=du/\alpha\) and changing the bounds to \(u(a)=\alpha(a-h)\) and \(u(b)=\alpha(b-h)\). If \(\alpha<0\), the transformed bounds are in reversed order; their orientation is part of the formula, so no extra sign change should be inserted by guesswork.
Difference Forms and Inverse Sine
For \(R>0\), the inverse-square-root form \(1/\sqrt{R^2-u^2}\) has antiderivative \(\arcsin(u/R)\) on \(-R<u<R\). The square-root form \(\sqrt{R^2-u^2}\) has antiderivative \(\tfrac12\bigl(u\sqrt{R^2-u^2}+R^2\arcsin(u/R)\bigr)\). The inverse sine, \(\arcsin\), takes values in \([-\pi/2,\pi/2]\); its argument must lie in \([-1,1]\).
Worked Example: A Square Root That Vanishes at an Endpoint
Evaluate \(\int_0^2\sqrt{4x-x^2}\,dx\).
First complete the square: \(4x-x^2=4-(x-2)^2\). On \(0\leq x\leq 2\), the radicand ranges from \(0\) to \(4\), so the square root is real throughout. Set \(u=x-2\); then \(du=dx\), and the bounds become \(u=-2\) and \(u=0\). The endpoint where the radicand is zero is allowed because the integrand is continuous there.
At \(u=0\), the antiderivative is \(0\). At \(u=-2\), \(\sqrt{4-(-2)^2}=0\) and \(\arcsin(-1)=-\pi/2\), so its value is \(\tfrac12(0+4(-\pi/2))=-\pi\). Therefore the definite integral is \(0-(-\pi)=\pi\).
Check the antiderivative: differentiating \(\tfrac12\bigl(u\sqrt{4-u^2}+4\arcsin(u/2)\bigr)\) gives \(\sqrt{4-u^2}\) for \(-2<u<2\). The endpoint evaluation uses the continuous extension of that antiderivative to \(u=-2\) and \(u=0\), confirming the result \(\pi\).
Worked Example: A Reciprocal Difference Form
Evaluate \(\int_0^1 \frac{dx}{\sqrt{9-(2x-1)^2}}\).
For \(0\leq x\leq 1\), \(2x-1\) ranges from \(-1\) to \(1\). Hence \(9-(2x-1)^2\geq 8>0\), and the integrand is continuous on the full interval. Set \(u=2x-1\), so \(du=2\,dx\), \(dx=du/2\), and the transformed bounds are \(-1\) and \(1\).
Substitution of the endpoints gives \(\tfrac12\bigl(\arcsin(1/3)-\arcsin(-1/3)\bigr)\). Since \(\arcsin\) is odd, this equals \(\arcsin(1/3)\).
Check by differentiating the antiderivative in the original variable: \(\frac{d}{dx}\bigl(\tfrac12\arcsin((2x-1)/3)\bigr)=1/\sqrt{9-(2x-1)^2}\). At \(x=1\) and \(x=0\), its values are respectively \(\tfrac12\arcsin(1/3)\) and \(-\tfrac12\arcsin(1/3)\), whose difference is the answer.
Positive Forms and Logarithmic Endpoint Values
For \(c>0\), the radicand \(u^2+c^2\) is positive for every real \(u\). Two useful antiderivatives are \(\ln\bigl(u+\sqrt{u^2+c^2}\bigr)\) for \(1/\sqrt{u^2+c^2}\), and \(\tfrac12\bigl(u\sqrt{u^2+c^2}+c^2\ln(u+\sqrt{u^2+c^2})\bigr)\) for \(\sqrt{u^2+c^2}\). The logarithm’s argument is positive because \(\sqrt{u^2+c^2}>|u|\), so \(u+\sqrt{u^2+c^2}>0\).
Worked Example: A Square Root of a Positive Completed Square
Evaluate \(\int_1^3\sqrt{(x-1)^2+4}\,dx\).
The radicand is at least \(4\), so the integrand is defined and continuous throughout the interval. Let \(u=x-1\). Then \(du=dx\), and \(x=1,3\) correspond to \(u=0,2\). Using the positive-form antiderivative with \(c=2\), we obtain
At \(u=2\), \(\sqrt{u^2+4}=\sqrt{8}=2\sqrt2\), so the antiderivative is \(2\sqrt2+2\ln(2+2\sqrt2)\). At \(u=0\), it is \(2\ln 2\). Subtracting gives \(2\sqrt2+2\ln(1+\sqrt2)\).
To check the formula, write \(s=\sqrt{u^2+4}\). The derivative of \(\ln(u+s)\) is \((1+u/s)/(u+s)=1/s\). The derivative of the displayed antiderivative is therefore \(\tfrac12(s+u^2/s+4/s)=s\), as required. The endpoint values above then verify the definite result.
Worked Example: A Reciprocal Positive Form
Evaluate \(\int_0^1 \frac{dx}{\sqrt{(x+1)^2+1}}\).
The radicand is positive on the whole interval. Set \(u=x+1\), so \(du=dx\); the bounds change from \(x=0,1\) to \(u=1,2\). Applying the logarithmic antiderivative with \(c=1\) gives
Both logarithm arguments are positive. Differentiating \(\ln(u+\sqrt{u^2+1})\) gives \(1/\sqrt{u^2+1}\), and since \(du/dx=1\), its derivative in \(x\) is the original integrand. Substitution of \(u=2\) and \(u=1\) gives exactly the stated endpoint difference.
A Common Bound-Transformation Error
In the reciprocal difference-form example, a tempting but incorrect step is to substitute \(u=2x-1\), write \(dx=du/2\), and keep the old bounds \(0\) and \(1\). That produces \(\tfrac12\int_0^1du/\sqrt{9-u^2}=\tfrac12\arcsin(1/3)\), which is not the original integral.
The error is that \(0\) and \(1\) were values of \(x\), not the corresponding values of \(u\). The correct transformed bounds are \(u=2(0)-1=-1\) and \(u=2(1)-1=1\). With those bounds, the integral is \(\tfrac12\bigl(\arcsin(1/3)-\arcsin(-1/3)\bigr)=\arcsin(1/3)\). The differential factor and the bounds must both be transformed.
Interpreting the Endpoint Expressions
Inverse-trigonometric and logarithmic antiderivatives can make the final answer look less familiar than the original integral. That does not make the answer approximate or ambiguous: the endpoint expressions are exact values. For example, \(\arcsin(1/3)\) is an exact real number, even if it is not a familiar angle such as \(\pi/6\). A numerical approximation is optional unless the problem requests one.
Before accepting an exact result, check that every inverse-sine input lies in \([-1,1]\), every logarithm argument is positive, and each endpoint value corresponds to the transformed variable at that endpoint. These checks often reveal an invalid interval or an incorrect bound before it affects the final evaluation.
Check Your Understanding
Use the domain, substitution, and endpoint checks developed here to answer these questions.
- On what interval is \(\sqrt{16-(x-2)^2}\) real?
- Under \(u=3x+1\), what are the transformed bounds when \(x\) runs from \(0\) to \(2\)?
- Why is it valid for the radicand in a square-root integrand to equal zero at an endpoint, while a reciprocal square root at that endpoint needs separate treatment?
- What is the exact value of \(\int_0^1 dx/\sqrt{4-x^2}\)?
- Why is \(u+\sqrt{u^2+c^2}\) positive for every real \(u\) when \(c>0\)?
