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Chapter 7 · Integrals

Exercise 7.4 · Q4: ∫dx9−25x2\int\frac{dx}{\sqrt{9-25x^2}}

Exercise 7.4Integrals of some particular functions · Section 7.4
The question

Integrate the function (Exercise 7.4, "Integrate the functions in Exercises 1 to 23"):

19−25x2\frac{1}{\sqrt{9 - 25x^2}}
Exam answer

Let I=∫dx9−25x2I = \displaystyle\int\frac{dx}{\sqrt{9 - 25x^2}}⁠.

Put 5x=t5x = t⁠. Then 5 dx=dt5\,dx = dt⁠.

I=15∫dt32−t2I = \frac15\int\frac{dt}{\sqrt{3^2 - t^2}}

Using ∫dta2−t2=sin⁡−1ta+C\displaystyle\int\frac{dt}{\sqrt{a^2 - t^2}} = \sin^{-1}\frac{t}{a} + C⁠,

I=15sin⁡−1t3+CI = \frac15\sin^{-1}\frac{t}{3} + C

∴ I=15sin⁡−15x3+C\therefore\ I = \dfrac15\sin^{-1}\dfrac{5x}{3} + C⁠.

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

The one idea this problem is built on

Under the root is "a constant squared minus a square": 9−25x2=32−(5x)29 - 25x^2 = 3^2 - (5x)^2⁠. That is the shape of formula 7.4 (5), ∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C⁠. The substitution t=5xt = 5x makes the match exact, at the cost of a factor 15\frac15⁠.

Given, to find, and the plan
Given
f(x)=19−25x2f(x) = \dfrac{1}{\sqrt{9 - 25x^2}}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Find where ff is defined. (2) Substitute t=5xt = 5x⁠. (3) Use formula 7.4 (5) with a=3a = 3⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1Formula 7.4 (5)

∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C⁠, for −a<x<a-a < x < a (a>0a > 0⁠). Lesson: integrals of particular functions

2Linear substitution

t=5xt = 5x gives dt=5 dxdt = 5\,dx⁠, so dx=15 dtdx = \frac15\,dt⁠. Basics: substitution (the chain rule backwards)

3Derivative of sin⁡−1\sin^{-1}

ddusin⁡−1u=11−u2\frac{d}{du}\sin^{-1}u = \frac{1}{\sqrt{1 - u^2}} for −1<u<1-1 < u < 1⁠. Basics: integration undoes differentiation

Solution

  1. Where is the integrand defined?

    We need 9−25x2>09 - 25x^2 > 0⁠, that is, x2<925x^2 < \frac{9}{25}⁠, that is,

    −35<x<35.-\frac35 < x < \frac35.
    The root is in a denominator, so the quantity under it must be strictly positive. The answer is only claimed on this interval; outside it the integrand does not exist as a real number.

  2. Substitute t=5xt = 5x

    Write 9−25x2=32−(5x)29 - 25x^2 = 3^2 - (5x)^2 and put t=5xt = 5x⁠, so dt=5 dxdt = 5\,dx⁠, dx=15 dtdx = \frac15\,dt⁠:

    ∫dx32−(5x)2=15∫dt32−t2.\int\frac{dx}{\sqrt{3^2 - (5x)^2}} = \frac15\int\frac{dt}{\sqrt{3^2 - t^2}}.

    For −35<x<35-\frac35 < x < \frac35⁠, the new variable satisfies −3<t<3-3 < t < 3⁠.

    Formula 7.4 (5) needs the variable squared with coefficient 1. Since 25x2=(5x)225x^2 = (5x)^2⁠, taking t=5xt = 5x achieves that. The range of tt is exactly where formula 7.4 (5) holds with a=3a = 3⁠.
    Integration by substitution, Section 7.3.1. Lesson: integration by substitution

  3. Use formula 7.4 (5) with a=3a = 3

    15∫dt32−t2=15sin⁡−1t3+C.\frac15\int\frac{dt}{\sqrt{3^2 - t^2}} = \frac15\sin^{-1}\frac t3 + C.
    "Constant squared minus a square, under a root" is formula 7.4 (5). The book proves it by putting x=asin⁡θx = a\sin\theta⁠. Taking −π2<θ<π2-\frac\pi2 < \theta < \frac\pi2 (the book does not state the range), cos⁡θ>0\cos\theta > 0⁠, so a2−a2sin⁡2θ=acos⁡θ\sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta with no sign trouble, and θ=sin⁡−1ta\theta = \sin^{-1}\frac ta is in the principal range.
    Formula 7.4 (5). Lesson: integrals of particular functions

  4. Return to xx

    ∫dx9−25x2=15sin⁡−15x3+C.\int\frac{dx}{\sqrt{9 - 25x^2}} = \frac15\sin^{-1}\frac{5x}{3} + C.

    On the domain, −1<5x3<1-1 < \frac{5x}{3} < 1⁠, so sin⁡−15x3\sin^{-1}\frac{5x}{3} is defined.

    Replace tt by 5x5x⁠. The same answer comes from taking 25 outside the root: 9−25x2=5(35)2−x2\sqrt{9 - 25x^2} = 5\sqrt{\left(\frac35\right)^2 - x^2}⁠, and formula 7.4 (5) with a=35a = \frac35 gives 15sin⁡−1x3/5\frac15\sin^{-1}\frac{x}{3/5}⁠, which is 15sin⁡−15x3\frac15\sin^{-1}\frac{5x}{3}⁠.

  5. Check: differentiate the answer back

    Let F(x)=15sin⁡−1uF(x) = \frac15\sin^{-1}u with u=5x3u = \frac{5x}{3}⁠, u′=53u' = \frac53⁠. Then

    F′(x)=15⋅11−25x29⋅53=13⋅39−25x2=19−25x2. ✓\begin{aligned} F'(x) &= \frac15\cdot\frac{1}{\sqrt{1 - \frac{25x^2}{9}}}\cdot\frac53 \\ &= \frac13\cdot\frac{3}{\sqrt{9 - 25x^2}} = \frac{1}{\sqrt{9 - 25x^2}}. \ \checkmark \end{aligned}
    Chain rule with outer sin⁡−1u\sin^{-1}u and inner u=5x3u = \frac{5x}{3}⁠. Then 1−25x29=9−25x23\sqrt{1 - \frac{25x^2}{9}} = \frac{\sqrt{9 - 25x^2}}{3} (take 19\frac19 out of the root as 13\frac13⁠), which turns 13⋅1…/3\frac13\cdot\frac{1}{\sqrt{\dots}/3} into the integrand.
    Standard formula (viii), Section 7.2. Basics: integration undoes differentiation
Result

∫dx9−25x2=15sin⁡−15x3+C,\int\frac{dx}{\sqrt{9 - 25x^2}} = \frac15\sin^{-1}\frac{5x}{3} + C, valid for −35<x<35-\frac35 < x < \frac35⁠. (Textbook answer: 15sin⁡−15x3+C\frac15\sin^{-1}\frac{5x}{3} + C⁠, the same.)

Check it with numbers

Take x=0.3x = 0.3 (inside the domain) and h=0.001h = 0.001⁠:

  • F(0.301)=0.10510487F(0.301) = 0.10510487 and F(0.299)=0.10433507F(0.299) = 0.10433507⁠,
  • slope ≈0.10510487−0.104335070.002=0.3849\approx \frac{0.10510487 - 0.10433507}{0.002} = 0.3849⁠,
  • integrand: 19−2.25=16.75=0.3849\frac{1}{\sqrt{9 - 2.25}} = \frac{1}{\sqrt{6.75}} = 0.3849⁠. ✓
Common mistakes
  1. Writing sin⁡−1x3\sin^{-1}\frac{x}{3} or sin⁡−15x9\sin^{-1}\frac{5x}{9}⁠. The variable is t=5xt = 5x and a=3a = 3 (not 9), so the argument is 5x3\frac{5x}{3}⁠.
  2. Losing the 15\frac15⁠. dx=15 dtdx = \frac15\,dt⁠.
  3. Using a logarithm formula. "Constant minus square" under a root gives sin⁡−1\sin^{-1}⁠; "square minus constant" or "square plus constant" gives a logarithm.
What to take away
  1. ∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + C⁠, on −a<x<a-a < x < a⁠.
  2. For p2−q2x2\sqrt{p^2 - q^2x^2}⁠: put t=qxt = qx⁠, use a=pa = p⁠, and divide by qq⁠.
  3. The domain of the integrand is exactly where the sin⁡−1\sin^{-1} answer is defined.

Every fact used, and where it comes from

FactUsed inSource
Substitution t=5xt = 5x⁠, dt=5 dxdt = 5\,dxStep 2Section 7.3.1 (Lesson: substitution)
∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + CSteps 3, 4Formula 7.4 (5), Section 7.4
Principal range of sin⁡−1\sin^{-1} is [−π2,π2]\left[-\frac\pi2, \frac\pi2\right]Step 3Chapter 2, Section 2.2
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}Step 5Standard formula (viii), Section 7.2
Chain rule; F′=fF' = f checks the answerStep 5Theorem 4, Section 5.3.1; Property (I), Section 7.2.1
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