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Chapter 7 · Integrals

Exercise 7.2 · Q26: ∫cos⁡xx dx\int\frac{\cos\sqrt{x}}{\sqrt{x}}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function

cos⁡xx.\frac{\cos\sqrt{x}}{\sqrt{x}}.
Exam answer

Let I=∫cos⁡xx dxI = \displaystyle\int\frac{\cos\sqrt x}{\sqrt x}\,dx⁠.

Put x=t\sqrt x = t⁠. Then dx2x=dt\frac{dx}{2\sqrt x} = dt⁠.

I=2∫cos⁡t dt=2sin⁡t+CI = 2\int\cos t\,dt = 2\sin t + C

∴ I=2sin⁡x+C\therefore\ I = 2\sin\sqrt x + C⁠.

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

The one idea this problem is built on

The derivative of x\sqrt{x} is 12x\frac{1}{2\sqrt{x}}⁠, and the integrand contains 1x\frac{1}{\sqrt{x}}⁠, which is twice that. So put t=xt = \sqrt{x}⁠: then dxx=2 dt\frac{dx}{\sqrt{x}} = 2\,dt⁠, and the integral becomes 2∫cos⁡t dt=2sin⁡t+C2\int\cos t\,dt = 2\sin t + C⁠.

Given, to find, and the plan
Given
f(x)=cos⁡xxf(x) = \dfrac{\cos\sqrt{x}}{\sqrt{x}}⁠, defined for x>0x > 0 (x\sqrt{x} needs x≥0x \ge 0⁠, and it is in the denominator, so x≠0x \ne 0⁠).
To find
∫cos⁡xx dx\displaystyle\int \frac{\cos\sqrt{x}}{\sqrt{x}}\,dx⁠.
Plan
(1) Find the derivative of x\sqrt{x}⁠. (2) Substitute t=xt = \sqrt{x} and rewrite dxx\frac{dx}{\sqrt{x}}⁠. (3) Integrate cos⁡t\cos t⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1The derivative of x\sqrt{x}

x=x1/2\sqrt{x} = x^{1/2}⁠, so ddxx=12x−1/2=12x\frac{d}{dx}\sqrt{x} = \frac12x^{-1/2} = \frac{1}{2\sqrt{x}} for x>0x > 0⁠. Basics: powers and the power rule

2Substitution

If t=g(x)t = g(x)⁠, then dt=g′(x) dxdt = g'(x)\,dx⁠; a constant multiple of g′(x) dxg'(x)\,dx can be traded for dtdt⁠. Basics: substitution

3The integral of cos⁡\cos

∫cos⁡t dt=sin⁡t+C\int\cos t\,dt = \sin t + C (standard formula (ii), Section 7.2). Basics: integration undoes differentiation

Solution

  1. Find the derivative of the inner function

    The cosine is applied to x\sqrt{x}⁠. Its derivative is

    ddxx=ddxx1/2=12x−1/2=12x.\frac{d}{dx}\sqrt{x} = \frac{d}{dx}x^{1/2} = \frac12x^{-1/2} = \frac{1}{2\sqrt{x}}.

    The integrand's factor 1x\frac{1}{\sqrt{x}} is 22 times this.

    The integrand has the shape cos⁡(g(x))⋅g′(x)\cos(g(x))\cdot g'(x) up to the constant 22⁠, with g(x)=xg(x) = \sqrt{x}⁠. This is the same situation as Example 5(iii) of the book, where t=xt = \sqrt{x} was also used.
    Power rule with n=12n = \frac12⁠. Basics: the power rule

  2. Substitute t=xt = \sqrt{x}

    Let t=xt = \sqrt{x}⁠, so t>0t > 0 for x>0x > 0⁠. By Step 1, dt=dx2xdt = \frac{dx}{2\sqrt{x}}⁠, so

    dxx=2 dt.\frac{dx}{\sqrt{x}} = 2\,dt.

    The integral becomes

    ∫cos⁡x⋅dxx=2∫cos⁡t dt.\int\cos\sqrt{x}\cdot\frac{dx}{\sqrt{x}} = 2\int\cos t\,dt.
    x\sqrt{x} is differentiable for every x>0x > 0⁠, which is the whole domain of the integrand, so the substitution is valid throughout. No xx is left after substituting.
    Change of variable, Section 7.3.1; Property (IV), Section 7.2.1. Lesson: integration by substitution

  3. Integrate and return to xx

    2∫cos⁡t dt=2sin⁡t+C=2sin⁡x+C.\begin{aligned} 2\int\cos t\,dt &= 2\sin t + C \\ &= 2\sin\sqrt{x} + C. \end{aligned}
    Standard formula (ii), Section 7.2.

  4. Check: differentiate the answer back

    By the chain rule (outer sin⁡u\sin u⁠, inner u=xu = \sqrt{x} with u′=12xu' = \frac{1}{2\sqrt{x}}⁠):

    ddx[2sin⁡x]=2cos⁡x⋅12x=cos⁡xx.\begin{aligned} \frac{d}{dx}\left[2\sin\sqrt{x}\right] &= 2\cos\sqrt{x}\cdot\frac{1}{2\sqrt{x}} \\ &= \frac{\cos\sqrt{x}}{\sqrt{x}}. \end{aligned}

    This is the integrand, so the answer is correct for x>0x > 0⁠.

    Chain rule, Theorem 4, Section 5.3.1; ddusin⁡u=cos⁡u\frac{d}{du}\sin u = \cos u⁠.
Result ∫cos⁡xx dx=2sin⁡x+C,x>0.\int \frac{\cos\sqrt{x}}{\sqrt{x}}\,dx = 2\sin\sqrt{x} + C, \qquad x > 0.

(Textbook answer: 2sin⁡x+C2\sin\sqrt{x} + C⁠.)

Check it with numbers

Let F(x)=2sin⁡xF(x) = 2\sin\sqrt{x} and look near x=1x = 1 with h=0.001h = 0.001⁠:

  • F(1.001)≈1.683481927F(1.001) \approx 1.683481927 and F(0.999)≈1.682401322F(0.999) \approx 1.682401322⁠.
  • Slope estimate: F(1.001)−F(0.999)0.002\dfrac{F(1.001) - F(0.999)}{0.002} ≈0.540302\approx 0.540302⁠.
  • Integrand at x=1x = 1⁠: cos⁡11≈0.540302\dfrac{\cos 1}{1} \approx 0.540302⁠.

They agree.

Common mistakes
  1. Writing sin⁡x\sin\sqrt{x} or 12sin⁡x\frac12\sin\sqrt{x}⁠. Since dx=2t dtdx = 2t\,dt⁠, the factor is 22⁠, not 11 or 12\frac12⁠. Differentiating sin⁡x\sin\sqrt{x} gives only half the integrand.
  2. Treating cos⁡x\cos\sqrt{x} as cos⁡x\sqrt{\cos x}⁠. The square root is inside the cosine.
  3. Forgetting the domain. The answer is for x>0x > 0 only.
What to take away
  1. 1x\frac{1}{\sqrt{x}} next to a function of x\sqrt{x} signals t=xt = \sqrt{x}⁠, with dxx=2 dt\frac{dx}{\sqrt{x}} = 2\,dt⁠.
  2. In general ∫f(x)x dx=2F(x)+C\int\frac{f(\sqrt{x})}{\sqrt{x}}\,dx = 2F(\sqrt{x}) + C⁠, where FF is an antiderivative of ff⁠.

Every fact used, and where it comes from

FactUsed inSource
ddxx=12x\frac{d}{dx}\sqrt{x} = \frac{1}{2\sqrt{x}}⁠, x>0x > 0Steps 1, 2, 4Power rule, Table 5.3; formula (i), Section 7.2
Change of variable (substitution)Step 2Section 7.3.1; Example 5(iii) (Lesson: integration by substitution)
∫kf=k∫f\int k f = k\int fStep 2Property (IV), Section 7.2.1
∫cos⁡t dt=sin⁡t+C\int\cos t\,dt = \sin t + CStep 3Standard formula (ii), Section 7.2
Chain ruleStep 4Theorem 4, Section 5.3.1
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