Class 12 Maths, Worked in Full Past papersGet updates
Chapter 7 · Integrals

Exercise 7.4 · Q25 (MCQ): ∫dx9x−4x2\int\frac{dx}{\sqrt{9x - 4x^2}}

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

Choose the correct answer.

∫dx9x−4x2  equals\int\frac{dx}{\sqrt{9x - 4x^2}} \ \text{ equals}

(A) 19sin⁡−1(9x−88)+C\frac19\sin^{-1}\left(\frac{9x - 8}{8}\right) + C    (B) 12sin⁡−1(8x−99)+C\frac12\sin^{-1}\left(\frac{8x - 9}{9}\right) + C
(C) 13sin⁡−1(9x−88)+C\frac13\sin^{-1}\left(\frac{9x - 8}{8}\right) + C    (D) 12sin⁡−1(9x−89)+C\frac12\sin^{-1}\left(\frac{9x - 8}{9}\right) + C

Exam answer

Here 9x−4x2=4[(98)2−(x−98)2]9x - 4x^2 = 4\left[\left(\frac98\right)^2 - \left(x - \frac98\right)^2\right]⁠. [Using ∫dxa2−x2\displaystyle\int\frac{dx}{\sqrt{a^2 - x^2}} =sin⁡−1xa+C= \sin^{-1}\frac xa + C]

∫dx9x−4x2=12∫dx(98)2−(x−98)2=12sin⁡−1x−9898+C=12sin⁡−1(8x−99)+C\begin{aligned} &\int\frac{dx}{\sqrt{9x - 4x^2}} \\ &= \frac12\int\frac{dx}{\sqrt{\left(\frac98\right)^2 - \left(x - \frac98\right)^2}} \\ &= \frac12\sin^{-1}\frac{x - \frac98}{\frac98} + C \\ &= \frac12\sin^{-1}\left(\frac{8x - 9}{9}\right) + C \end{aligned}

Answer: (B) 12sin⁡−1(8x−99)+C\dfrac12\sin^{-1}\left(\dfrac{8x - 9}{9}\right) + C

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

The one idea this problem is built on

Take the 4 out of the quadratic and complete the square:

9x−4x2=4[(98)2−(x−98)2].9x - 4x^2 = 4\left[\left(\frac98\right)^2 - \left(x - \frac98\right)^2\right].

So 9x−4x2=2a2−t2\sqrt{9x - 4x^2} = 2\sqrt{a^2 - t^2} with a=98a = \frac98⁠, t=x−98t = x - \frac98⁠, and formula 7.4 (5) gives 12sin⁡−1ta=12sin⁡−18x−99\frac12\sin^{-1}\frac ta = \frac12\sin^{-1}\frac{8x - 9}{9}⁠: option (B).

Given, to find, and the plan
Given
The integrand 19x−4x2\frac{1}{\sqrt{9x - 4x^2}} and four options.
To find
Which option is ∫dx9x−4x2\int\frac{dx}{\sqrt{9x - 4x^2}}⁠.
Plan
(1) Find the domain. (2) Take out 4 and complete the square. (3) Shift and use 7.4 (5). (4) Differentiate back. (5) Test every option.

Basics you need

1Completing the square

x2−px=(x−p2)2−p24x^2 - px = \left(x - \frac p2\right)^2 - \frac{p^2}{4}⁠, after making the x2x^2 coefficient 1. Basics: completing the square

2Formula 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. Lesson: integrals of particular functions

3Testing an option

An option F+CF + C is right exactly when F′F' equals the integrand at every point of the domain. Basics: integration undoes differentiation

Solution

  1. Where the integrand exists

    We need 9x−4x2=x(9−4x)>09x - 4x^2 = x(9 - 4x) > 0⁠, that is 0<x<940 < x < \frac94⁠.

    The quadratic is under a square root in a denominator, so it must be strictly positive. The product x(9−4x)x(9 - 4x) is positive exactly between its roots 0 and 94\frac94⁠.

  2. Take out the 4 and complete the square

    9x−4x2=−4(x2−94x)=−4[(x−98)2−8164]=4[(98)2−(x−98)2].\begin{aligned} 9x - 4x^2 &= -4\left(x^2 - \frac94 x\right) \\ &= -4\left[\left(x - \frac98\right)^2 - \frac{81}{64}\right] \\ &= 4\left[\left(\frac98\right)^2 - \left(x - \frac98\right)^2\right]. \end{aligned}

    So 9x−4x2=2(98)2−(x−98)2\sqrt{9x - 4x^2} = 2\sqrt{\left(\frac98\right)^2 - \left(x - \frac98\right)^2}⁠.

    Taking out −4-4 makes the x2x^2 coefficient 1. Half of −94-\frac94 is −98-\frac98⁠, and (98)2=8164\left(\frac98\right)^2 = \frac{81}{64}⁠. The square root of a product of positive numbers is the product of the roots, and 4=2\sqrt4 = 2⁠.
    Type 7.4 (8). Basics: completing the square

  3. Shift and use 7.4 (5)

    Put t=x−98t = x - \frac98⁠, dt=dxdt = dx⁠, and a=98a = \frac98⁠. On the domain, −98<t<98-\frac98 < t < \frac98⁠. Then

    ∫dx9x−4x2=12∫dta2−t2=12sin⁡−1ta+C=12sin⁡−1x−9898+C=12sin⁡−18x−99+C.\begin{aligned} \int\frac{dx}{\sqrt{9x - 4x^2}} &= \frac12\int\frac{dt}{\sqrt{a^2 - t^2}} = \frac12\sin^{-1}\frac ta + C \\ &= \frac12\sin^{-1}\frac{x - \frac98}{\frac98} + C = \frac12\sin^{-1}\frac{8x - 9}{9} + C. \end{aligned}
    The domain 0<x<940 < x < \frac94 is exactly ∣t∣<98|t| < \frac98⁠, where 7.4 (5) holds. Multiplying the top and bottom of x−9/89/8\frac{x - 9/8}{9/8} by 8 gives 8x−99\frac{8x - 9}{9}⁠.
    Formula 7.4 (5); substitution, Section 7.3.1. Lesson: integrals of particular functions

  4. Check: differentiate the answer

    Chain rule with inner function u=8x−99u = \frac{8x - 9}{9}⁠, u′=89u' = \frac89⁠:

    ddx[12sin⁡−1u]=12⋅891−(8x−9)281=481−(8x−9)2=4144x−64x2=449x−4x2=19x−4x2. ✓\begin{aligned} \frac{d}{dx}\left[\frac12\sin^{-1}u\right] &= \frac12\cdot\frac{\frac89}{\sqrt{1 - \frac{(8x - 9)^2}{81}}} \\ &= \frac{4}{\sqrt{81 - (8x - 9)^2}} = \frac{4}{\sqrt{144x - 64x^2}} \\ &= \frac{4}{4\sqrt{9x - 4x^2}} = \frac{1}{\sqrt{9x - 4x^2}}. \ \checkmark \end{aligned}
    12⋅89=49\frac12\cdot\frac89 = \frac49⁠, and multiplying the top and bottom by 9 gives 481−(8x−9)2\frac{4}{\sqrt{81 - (8x - 9)^2}}⁠. Then (8x−9)2=64x2−144x+81(8x - 9)^2 = 64x^2 - 144x + 81⁠, so 81−(8x−9)2=144x−64x281 - (8x - 9)^2 = 144x - 64x^2 =16(9x−4x2)= 16(9x - 4x^2)⁠, and 16=4\sqrt{16} = 4⁠.
    Derivative (viii), Section 7.2; chain rule (Theorem 4, Section 5.3.1). Basics: integration undoes differentiation

  5. Examine every option

    Differentiate each option exactly as in Step 4, and compare at x=1x = 1⁠, where the integrand is 15≈0.4472\frac{1}{\sqrt5} \approx 0.4472⁠.

    • (A) With u=9x−88u = \frac{9x - 8}{8}⁠, u′=98u' = \frac98⁠: 1−u2=1864−(9x−8)2\sqrt{1 - u^2} = \frac18\sqrt{64 - (9x - 8)^2}⁠, and 64−(9x−8)2=144x−81x264 - (9x - 8)^2 = 144x - 81x^2 =9(16x−9x2)= 9(16x - 9x^2)⁠. So ddxsin⁡−1u=9316x−9x2\frac{d}{dx}\sin^{-1}u = \frac{9}{3\sqrt{16x - 9x^2}} =316x−9x2= \frac{3}{\sqrt{16x - 9x^2}}⁠, and option (A) has derivative 1316x−9x2\frac{1}{3\sqrt{16x - 9x^2}}⁠. At x=1x = 1⁠: 137≈0.1260\frac{1}{3\sqrt7} \approx 0.1260⁠. Wrong.
    • (B) Derivative 19x−4x2\frac{1}{\sqrt{9x - 4x^2}} for every xx in the domain (Step 4). At x=1x = 1⁠: 0.44720.4472⁠. ✓ Correct.
    • (C) By the working in (A), the derivative is 13⋅316x−9x2\frac13\cdot\frac{3}{\sqrt{16x - 9x^2}} =116x−9x2= \frac{1}{\sqrt{16x - 9x^2}}⁠. At x=1x = 1⁠: 17≈0.3780\frac{1}{\sqrt7} \approx 0.3780⁠. Wrong: it is the integral of 116x−9x2\frac{1}{\sqrt{16x - 9x^2}}⁠, a different function.
    • (D) With u=9x−89u = \frac{9x - 8}{9}⁠, u′=1u' = 1⁠: 1−u2=1981−(9x−8)2\sqrt{1 - u^2} = \frac19\sqrt{81 - (9x - 8)^2} and 81−(9x−8)281 - (9x - 8)^2 =17+144x−81x2= 17 + 144x - 81x^2⁠, so the derivative is 9217+144x−81x2\frac{9}{2\sqrt{17 + 144x - 81x^2}}⁠. At x=1x = 1⁠: 985≈0.5031\frac{9}{8\sqrt5} \approx 0.5031⁠. Wrong.
    A single point where the derivative differs from the integrand is enough to reject an option. In (A) and (C) the derivative of sin⁡−19x−88\sin^{-1}\frac{9x - 8}{8} is 316x−9x2\frac{3}{\sqrt{16x - 9x^2}}⁠, so (A) gives 19\frac19 of that and (C) gives 13\frac13 of it; neither involves 9x−4x29x - 4x^2⁠.
Result ∫dx9x−4x2=12sin⁡−1(8x−99)+C,\int\frac{dx}{\sqrt{9x - 4x^2}} = \frac12\sin^{-1}\left(\frac{8x - 9}{9}\right) + C,

valid for 0<x<940 < x < \frac94⁠. The correct option is (B). (Textbook answer: B.)

Check it with numbers

By Section 7.8, ∫12dx9x−4x2=F(2)−F(1)\int_1^2 \frac{dx}{\sqrt{9x - 4x^2}} = F(2) - F(1) with F(x)=12sin⁡−18x−99F(x) = \frac12\sin^{-1}\frac{8x - 9}{9}⁠:

  • a computer's sum of very many thin strips (found without our answer) gives 0.50123180.5012318⁠;
  • F(2)−F(1)F(2) - F(1) =12[sin⁡−179−sin⁡−1(−19)]= \frac12\left[\sin^{-1}\frac79 - \sin^{-1}\left(-\frac19\right)\right] =0.5012318= 0.5012318⁠. ✓
Common mistakes
  1. Forgetting the 12\frac12 from 4\sqrt4⁠. Taking 4 out from under the root gives a factor 12\frac12 in front of the integral.
  2. Mixing up the coefficients. Options (A) and (C) use 9x−88\frac{9x - 8}{8}⁠, which belongs to the quadratic 16x−9x216x - 9x^2⁠; option (D) uses 9x−89\frac{9x - 8}{9}⁠, which belongs to 17+144x−81x217 + 144x - 81x^2⁠. Neither is 9x−4x29x - 4x^2⁠.
  3. Completing the square with 94\frac94 instead of 98\frac98⁠. The shift is half the coefficient of xx after dividing by 4.
What to take away
  1. For ∫dxc+bx−ax2\int\frac{dx}{\sqrt{c + bx - ax^2}}⁠, first take aa out (giving 1a\frac{1}{\sqrt a} in front), then complete the square.
  2. For MCQs, differentiate each option and compare with the integrand at a convenient point.

Every fact used, and where it comes from

FactUsed inSource
Completing the squareStep 2Type 7.4 (8) (Basics: completing the square)
∫dxa2−x2=sin⁡−1xa+C\int\frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}\frac xa + CStep 3Formula 7.4 (5)
Substitution t=x−98t = x - \frac98Step 3Section 7.3.1 (Basics: substitution)
ddxsin⁡−1x=11−x2\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1 - x^2}}⁠; chain ruleSteps 4, 5Derivative (viii), Section 7.2; Theorem 4, Section 5.3.1
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a)Numerical checkSection 7.8 (Lesson: definite integrals)
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →