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Chapter 7 · Integrals

Miscellaneous · Q3: ∫dxxax−x2\int\frac{dx}{x\sqrt{ax-x^2}}

Miscellaneous ExerciseIntegration by substitution (x = a/t) · Miscellaneous Exercise on Chapter 7
The question

Integrate the functions in Exercises 1 to 23.

3. 1xax−x2\dfrac{1}{x\sqrt{ax-x^2}}   [Hint: Put x=atx = \dfrac at]

Exam answer

Let I=∫dxxax−x2I = \displaystyle\int\frac{dx}{x\sqrt{ax - x^2}}⁠, where a>0a > 0⁠, 0<x<a0 < x < a⁠.

Put x=atx = \frac{a}{t}⁠. Then dx=−at2 dtdx = -\frac{a}{t^2}\,dt⁠, and t>1t > 1⁠.

ax−x2=a2t−a2t2=a2(t−1)t2\begin{aligned} ax - x^2 &= \frac{a^2}{t} - \frac{a^2}{t^2} \\ &= \frac{a^2(t - 1)}{t^2} \end{aligned}

So xax−x2=a2t−1t2x\sqrt{ax - x^2} = \frac{a^2\sqrt{t - 1}}{t^2}⁠.

∴ I=∫t2a2t−1(−at2)dt=−1a∫(t−1)−1/2 dt=−2at−1+C\begin{aligned} \therefore\ I &= \int\frac{t^2}{a^2\sqrt{t - 1}}\left(-\frac{a}{t^2}\right)dt \\ &= -\frac1a\int(t - 1)^{-1/2}\,dt \\ &= -\frac2a\sqrt{t - 1} + C \end{aligned}

Since t=axt = \frac{a}{x}⁠, t−1=a−xxt - 1 = \frac{a - x}{x}⁠.

∴ I=−2aa−xx+C\therefore\ I = -\dfrac2a\sqrt{\dfrac{a - x}{x}} + C (for a>0a > 0⁠).

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The factor xx outside the root is what makes this hard. The reciprocal substitution x=atx = \frac at turns 1x\frac1x into ta\frac ta⁠, and after simplifying, all the powers of tt cancel except a single 1t−1\frac{1}{\sqrt{t-1}}⁠, which the power rule integrates at once.

Given, to find, and the plan
Given
f(x)=1xax−x2f(x) = \dfrac{1}{x\sqrt{ax-x^2}}⁠, with aa a non-zero constant. We take a>0a > 0⁠, the case the book's answer is written for; Step 6 treats a<0a < 0⁠.
To find
∫dxxax−x2\displaystyle\int\frac{dx}{x\sqrt{ax-x^2}}⁠.
Plan
(1) Find the domain. (2) Substitute x=atx = \frac at and rewrite every part. (3) Integrate in tt⁠. (4) Return to xx⁠. (5) Differentiate back. (6) The case a<0a < 0⁠.

Basics you need

1Substitution x=g(t)x = g(t)

If x=g(t)x = g(t) with gg differentiable and one-one on the interval, then dx=g′(t) dtdx = g'(t)\,dt and ∫f(x) dx=∫f(g(t)) g′(t) dt\int f(x)\,dx = \int f(g(t))\,g'(t)\,dt (Section 7.3.1). Basics: substitution

2Power rule

∫u−1/2 du=u1/21/2=2u+C\int u^{-1/2}\,du = \frac{u^{1/2}}{1/2} = 2\sqrt u + C (standard formula (i) with n=−12n = -\frac12⁠). Basics: the power rule

3p2=∣p∣\sqrt{p^2} = |p|

A square root is never negative, so a2=a\sqrt{a^2} = a only when a≥0a \ge 0⁠; for a<0a < 0 it is −a-a⁠. This decides the sign of the answer.

Solution

  1. Domain, and why the hint fits

    The root needs ax−x2=x(a−x)>0ax - x^2 = x(a - x) > 0 (strictly, because it is in a denominator), and x≠0x \ne 0⁠. With a>0a > 0⁠, the product x(a−x)x(a-x) is positive exactly when 0<x<a0 < x < a⁠. So we work on the interval 0<x<a0 < x < a⁠.

    The integrand is not a standard form: the outside factor xx blocks formula (5) of Section 7.4, and t=ax−x2t = ax - x^2 does not work because the numerator is not a multiple of its derivative a−2xa - 2x⁠. Putting x=atx = \frac at replaces the troublesome 1x\frac1x by ta\frac ta⁠, and the root simplifies as shown next. On 0<x<a0 < x < a the map t=axt = \frac ax is one-one and differentiable, and it sends the interval to t>1t > 1 (since x<ax < a means ax>1\frac ax > 1⁠).
    Substitution needs a one-one, differentiable change of variable on the interval (Basic 1).

  2. Substitute x=atx = \frac at⁠: rewrite dxdx and the root

    Differentiate: dx=−at2 dtdx = -\dfrac{a}{t^2}\,dt⁠. Inside the root:

    ax−x2=a⋅at−a2t2=a2t−a2t2=a2(t−1)t2.\begin{aligned} ax - x^2 &= a\cdot\frac at - \frac{a^2}{t^2} = \frac{a^2}{t} - \frac{a^2}{t^2} \\ &= \frac{a^2(t-1)}{t^2}. \end{aligned}

    Since a>0a > 0 and t>1t > 1⁠: ax−x2=at−1t\sqrt{ax - x^2} = \dfrac{a\sqrt{t-1}}{t}⁠. Then

    xax−x2=at⋅at−1t=a2t−1t2.x\sqrt{ax-x^2} = \frac at\cdot\frac{a\sqrt{t-1}}{t} = \frac{a^2\sqrt{t-1}}{t^2}.
    To take the square root of a2(t−1)t2\frac{a^2(t-1)}{t^2} we need a2=a\sqrt{a^2} = a and t2=t\sqrt{t^2} = t⁠, which hold because a>0a > 0 and t>0t > 0 (Basic 3). Also t−1>0t - 1 > 0⁠, so t−1\sqrt{t-1} is real and non-zero.
    ddt(at)=−at2\frac{d}{dt}\big(\frac at\big) = -\frac{a}{t^2} (power rule for derivatives, Chapter 5).

  3. Simplify the integral in tt and integrate

    ∫dxxax−x2=∫t2a2t−1⋅(−at2)dt=−1a∫(t−1)−1/2 dt=−1a⋅2t−1+C=−2at−1+C.\begin{aligned} \int\frac{dx}{x\sqrt{ax-x^2}} &= \int\frac{t^2}{a^2\sqrt{t-1}}\cdot\Big(-\frac{a}{t^2}\Big)dt \\ &= -\frac1a\int (t-1)^{-1/2}\,dt \\ &= -\frac1a\cdot 2\sqrt{t-1} + C \\ &= -\frac2a\sqrt{t-1} + C. \end{aligned}
    The t2t^2 from the reciprocal of xax−x2x\sqrt{ax-x^2} cancels the t2t^2 in dxdx⁠: that cancellation is why the hint works. What is left is a power of t−1t - 1⁠, integrated by the power rule with n=−12n = -\frac12 (and d(t−1)=dtd(t-1) = dt⁠).
    Standard formula (i), Section 7.2, with n=−12n = -\frac12⁠. Basics: the power rule

  4. Return to xx

    From x=atx = \frac at we have t=axt = \frac ax⁠, so t−1=ax−1=a−xxt - 1 = \frac ax - 1 = \dfrac{a - x}{x}⁠. Therefore

    ∫dxxax−x2=−2aa−xx+C.\int\frac{dx}{x\sqrt{ax-x^2}} = -\frac2a\sqrt{\frac{a-x}{x}} + C.
    An indefinite integral must be given in the original variable. This is exactly the textbook's answer.
    Substitution, Section 7.3.1. Lesson: integration by substitution

  5. Check: differentiate the answer

    Write g=a−xx=ax−1g = \frac{a-x}{x} = \frac ax - 1⁠, so g′=−ax2g' = -\frac{a}{x^2}⁠. By the chain rule:

    ddx(−2ag)=−2a⋅g′2g=−2a⋅−a/x22g=1x2(a−x)/x.\begin{aligned} \frac{d}{dx}\Big(-\frac2a\sqrt g\Big) &= -\frac2a\cdot\frac{g'}{2\sqrt g} = -\frac2a\cdot\frac{-a/x^2}{2\sqrt g} \\ &= \frac{1}{x^2\sqrt{(a-x)/x}}. \end{aligned}

    For x>0x > 0⁠: x2a−xx=xx2⋅a−xx=xax−x2x^2\sqrt{\frac{a-x}{x}} = x\sqrt{x^2\cdot\frac{a-x}{x}} = x\sqrt{ax - x^2}⁠. So the derivative is 1xax−x2\dfrac{1}{x\sqrt{ax-x^2}}⁠. ✓

    Moving one factor xx inside the root as x2x^2 is allowed only because x>0x > 0 here (x=x2x = \sqrt{x^2} for positive xx⁠). The derivative returns the integrand on the whole interval 0<x<a0 < x < a⁠.
    Chain rule; ddxg=g′2g\frac{d}{dx}\sqrt g = \frac{g'}{2\sqrt g}⁠, Chapter 5. Basics: integration undoes differentiation

  6. The case a<0a < 0

    Now x(a−x)>0x(a - x) > 0 means a<x<0a < x < 0⁠. The same substitution still gives t=ax>1t = \frac ax > 1⁠, but now a2=−a\sqrt{a^2} = -a⁠, so ax−x2=−at−1t\sqrt{ax-x^2} = \frac{-a\sqrt{t-1}}{t} and every sign after Step 2 flips:

    ∫dxxax−x2=2aa−xx+C(a<0).\int\frac{dx}{x\sqrt{ax-x^2}} = \frac2a\sqrt{\frac{a-x}{x}} + C \quad (a < 0).

    Both cases are covered by the single formula −2ax−x2ax+C-\dfrac{2\sqrt{ax-x^2}}{ax} + C⁠, because ax−x2x=a−xx\frac{\sqrt{ax-x^2}}{x} = \sqrt{\frac{a-x}{x}} when x>0x > 0 and −a−xx-\sqrt{\frac{a-x}{x}} when x<0x < 0⁠.

    The question does not say a>0a > 0⁠, so this case must be settled. The book's form −2aa−xx-\frac2a\sqrt{\frac{a-x}{x}} is correct for a>0a > 0⁠; for a<0a < 0 its derivative is the negative of the integrand (in Step 5 we would need x2⋅=−xax−x2x^2\sqrt{\cdot} = -x\sqrt{ax-x^2}⁠, as now x<0x < 0⁠).
    p2=∣p∣\sqrt{p^2} = |p| (Basic 3).
Result

For a>0a > 0 and 0<x<a0 < x < a⁠:

∫dxxax−x2=−2aa−xx+C.\int\frac{dx}{x\sqrt{ax-x^2}} = -\frac2a\sqrt{\frac{a-x}{x}} + C.

(Textbook answer: −2aa−xx+C-\frac2a\sqrt{\frac{a-x}{x}} + C⁠, the same.) For a<0a < 0 (a<x<0a < x < 0⁠) the answer is 2aa−xx+C\frac2a\sqrt{\frac{a-x}{x}} + C⁠; both cases are −2ax−x2ax+C-\frac{2\sqrt{ax-x^2}}{ax} + C⁠.

Check it with numbers

Take a=2a = 2 and integrate from x=12x = \frac12 to x=32x = \frac32 (inside 0<x<20 < x < 2⁠). With F(x)=−2−xxF(x) = -\sqrt{\frac{2-x}{x}}⁠:

F(32)−F(12)=−13+3=23≈1.1547005.\begin{aligned} F\big(\tfrac32\big) - F\big(\tfrac12\big) &= -\sqrt{\tfrac13} + \sqrt3 \\ &= \tfrac{2}{\sqrt3} \approx 1.1547005. \end{aligned}

Numerical integration from 12\frac12 to 32\frac32 also gives 1.15470051.1547005⁠. In tt the limits are t=4t = 4 and t=43t = \frac43⁠, and −1a∫44/3dtt−1=3−13-\frac1a\int_4^{4/3}\frac{dt}{\sqrt{t-1}} = \sqrt3 - \sqrt{\tfrac13} as well.

Common mistakes
  1. Losing the minus sign of dx=−at2 dtdx = -\frac{a}{t^2}\,dt⁠. The whole answer then has the wrong sign.
  2. Writing a2=a\sqrt{a^2} = a without checking the sign of aa⁠. That is what changes the answer when a<0a < 0⁠.
  3. Leaving the answer in tt⁠. Replace tt by ax\frac ax at the end.
What to take away
  1. When a power of xx multiplies a root of a quadratic in the denominator, try the reciprocal substitution x=atx = \frac at (or x=1tx = \frac1t⁠).
  2. Every time you take p2\sqrt{p^2}⁠, decide the sign of pp on your interval.
  3. Check by differentiating: it also tells you on which interval your formula is valid.

Every fact used, and where it comes from

FactUsed inSource
Sign of the product x(a−x)x(a-x)Steps 1, 6Algebra
Substitution x=g(t)x = g(t)⁠, dx=g′(t) dtdx = g'(t)\,dtSteps 1 to 4Section 7.3.1 (hint in the question)
p2=∣p∣\sqrt{p^2} = |p|Steps 2, 6Algebra
∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C⁠, n=−12n = -\frac12Step 3Standard formula (i), Section 7.2
Chain rule; derivative of g\sqrt gStep 5Chapter 5
All decimalsCheckRecomputed in the site's verification script
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