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

Exercise 7.3 · Q21: ∫sin⁡−1(cos⁡x) dx\int\sin^{-1}(\cos x)\,dx

Exercise 7.3Integration using trigonometric identities · Section 7.3.2
The question

Find the integral of the function

sin⁡−1(cos⁡x).\sin^{-1}(\cos x).
Exam answer

Let I=∫sin⁡−1(cos⁡x) dxI = \displaystyle\int\sin^{-1}(\cos x)\,dx⁠.

Using cos⁡x=sin⁡(π2−x)\cos x = \sin\left(\frac{\pi}{2} - x\right)⁠,

sin⁡−1(cos⁡x)=sin⁡−1(sin⁡(π2−x))\begin{aligned} &\sin^{-1}(\cos x) \\ &= \sin^{-1}\left(\sin\left(\frac{\pi}{2} - x\right)\right) \end{aligned}

For 0≤x≤π0 \le x \le \pi⁠, −π2≤π2−x≤π2-\frac{\pi}{2} \le \frac{\pi}{2} - x \le \frac{\pi}{2}⁠, so

sin⁡−1(cos⁡x)=π2−x\sin^{-1}(\cos x) = \frac{\pi}{2} - x ∴ I=∫(π2−x)dx\therefore\ I = \int\left(\frac{\pi}{2} - x\right)dx

∴ I=πx2−x22+C\therefore\ I = \dfrac{\pi x}{2} - \dfrac{x^2}{2} + C⁠, for 0≤x≤π0 \le x \le \pi⁠.

The answer key prints the same form, which holds for 0≤x≤π0 \le x \le \pi⁠; the book does not state the interval.

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

The one idea this problem is built on

Rewrite the inside as a sine: cos⁡x=sin⁡(π2−x)\cos x = \sin\left(\frac{\pi}{2} - x\right)⁠. Then sin⁡−1\sin^{-1} undoes sin⁡\sin⁠, but only when the angle lies in the principal range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. The angle π2−x\frac{\pi}{2} - x lies there exactly when 0≤x≤π0 \le x \le \pi⁠. On that interval the integrand is simply π2−x\frac{\pi}{2} - x⁠, a straight line, and its integral is πx2−x22+C\frac{\pi x}{2} - \frac{x^2}{2} + C⁠.

Given, to find, and the plan
Given
f(x)=sin⁡−1(cos⁡x)f(x) = \sin^{-1}(\cos x)⁠. Since −1≤cos⁡x≤1-1 \le \cos x \le 1 for every real xx⁠, ff is defined for all real xx⁠, with values in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.
To find
∫sin⁡−1(cos⁡x) dx\displaystyle\int \sin^{-1}(\cos x)\,dx⁠.
Plan
(1) Check the domain. (2) Write cos⁡x\cos x as a sine. (3) Find where sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta applies. (4) Integrate the resulting line on [0,π][0, \pi]⁠. (5) Differentiate back. (6) See what happens outside [0,π][0, \pi]⁠.

Basics you need

1The principal value branch of sin⁡−1\sin^{-1}

sin⁡−1\sin^{-1} has domain [−1,1][-1, 1] and range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] (Chapter 2, Section 2.2). So sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta only for θ\theta in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] (Section 2.3). Basics: trigonometric identities for integrals

2Complementary angles

sin⁡(π2−x)=cos⁡x\sin\left(\frac{\pi}{2} - x\right) = \cos x for every real xx⁠.

3Integrating a polynomial

∫k dx=kx+C\int k\,dx = kx + C and ∫x dx=x22+C\int x\,dx = \frac{x^2}{2} + C (standard formula (i), Section 7.2). Basics: powers and the power rule

Solution

  1. Where is the integrand defined?

    sin⁡−1\sin^{-1} accepts inputs in [−1,1][-1, 1]⁠, and cos⁡x\cos x always lies in [−1,1][-1, 1]⁠. So sin⁡−1(cos⁡x)\sin^{-1}(\cos x) is defined for every real xx⁠, and its value always lies in the principal range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

    Unlike the other questions in this exercise, nothing blows up here. The difficulty is different: sin⁡−1(cos⁡x)\sin^{-1}(\cos x) is not given by one formula for all xx⁠, as the next two steps show.
    Domain and range of sin⁡−1\sin^{-1}⁠: Chapter 2, Section 2.2.

  2. Write cos⁡x\cos x as a sine

    Since cos⁡x=sin⁡(π2−x)\cos x = \sin\left(\frac{\pi}{2} - x\right)⁠,

    sin⁡−1(cos⁡x)=sin⁡−1(sin⁡(π2−x)).\sin^{-1}(\cos x) = \sin^{-1}\left(\sin\left(\tfrac{\pi}{2} - x\right)\right).
    sin⁡−1\sin^{-1} is the inverse of sin⁡\sin⁠, not of cos⁡\cos⁠. To let it "undo" something, the inside must be a sine. The complementary-angle identity does exactly that.
    sin⁡(π2−θ)=cos⁡θ\sin\left(\frac{\pi}{2} - \theta\right) = \cos\theta (Class XI trigonometry). Basics: trigonometric identities

  3. Where does sin⁡−1\sin^{-1} undo sin⁡\sin⁠?

    sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta holds when −π2≤θ≤π2-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}⁠. With θ=π2−x\theta = \frac{\pi}{2} - x⁠:

    −π2≤π2−x≤π2  ⟺  −π≤−x≤0  ⟺  0≤x≤π.\begin{aligned} -\frac{\pi}{2} \le \frac{\pi}{2} - x \le \frac{\pi}{2} &\iff -\pi \le -x \le 0 \\ &\iff 0 \le x \le \pi. \end{aligned}

    Therefore

    sin⁡−1(cos⁡x)=π2−xfor 0≤x≤π.\sin^{-1}(\cos x) = \frac{\pi}{2} - x \quad\text{for } 0 \le x \le \pi.

    Spot checks: at x=0x = 0⁠, sin⁡−1(1)=π2\sin^{-1}(1) = \frac{\pi}{2}⁠; at x=π3x = \frac{\pi}{3}⁠, sin⁡−1(12)=π6=π2−π3\sin^{-1}\left(\frac12\right) = \frac{\pi}{6} = \frac{\pi}{2} - \frac{\pi}{3}⁠; at x=πx = \pi⁠, sin⁡−1(−1)=−π2\sin^{-1}(-1) = -\frac{\pi}{2}⁠.

    Subtracting π2\frac{\pi}{2} from all three parts and then multiplying by −1-1 (which reverses the inequalities) gives 0≤x≤π0 \le x \le \pi⁠. Outside this interval the angle π2−x\frac{\pi}{2} - x is not in the principal range, and sin⁡−1\sin^{-1} returns a different angle with the same sine (Step 6).
    sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x for x∈[−π2,π2]x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠: Chapter 2, Section 2.3.

  4. Integrate on [0,π][0, \pi]

    ∫sin⁡−1(cos⁡x) dx=∫(π2−x)dx=π2 x−x22+C=πx2−x22+C.\begin{aligned} \int \sin^{-1}(\cos x)\,dx &= \int\left(\frac{\pi}{2} - x\right)dx \\ &= \frac{\pi}{2}\,x - \frac{x^2}{2} + C \\ &= \frac{\pi x}{2} - \frac{x^2}{2} + C. \end{aligned}
    π2\frac{\pi}{2} is a constant, so its integral is π2x\frac{\pi}{2}x⁠; the power rule gives ∫x dx=x22\int x\,dx = \frac{x^2}{2}⁠. Property (V) lets us integrate the two terms separately.
    Standard formula (i), Section 7.2; Property (V), Section 7.2.1. Lesson: integration as anti-differentiation

  5. Check: differentiate the answer back

    ddx[πx2−x22]=π2−x=sin⁡−1(cos⁡x)(0≤x≤π).\begin{aligned} \frac{d}{dx}\left[\frac{\pi x}{2} - \frac{x^2}{2}\right] &= \frac{\pi}{2} - x \\ &= \sin^{-1}(\cos x) \quad (0 \le x \le \pi). \end{aligned}

    The second line is Step 3. So the answer is correct on [0,π][0, \pi] (with one-sided derivatives at the end points).

    Property (I), Section 7.2.1. Basics: integration undoes differentiation

  6. Outside [0,π][0, \pi] the formula changes

    Try x=−1x = -1⁠: cos⁡(−1)=cos⁡1\cos(-1) = \cos 1⁠, so sin⁡−1(cos⁡(−1))=π2−1≈0.570796\sin^{-1}(\cos(-1)) = \frac{\pi}{2} - 1 \approx 0.570796⁠, whereas π2−(−1)≈2.570796\frac{\pi}{2} - (-1) \approx 2.570796⁠. The line π2−x\frac{\pi}{2} - x is wrong here, and it would even leave the range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠.

    For −π≤x≤0-\pi \le x \le 0 use instead cos⁡x=sin⁡(π2+x)\cos x = \sin\left(\frac{\pi}{2} + x\right)⁠, whose angle π2+x\frac{\pi}{2} + x lies in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠. So there

    sin⁡−1(cos⁡x)=π2+x,∫sin⁡−1(cos⁡x) dx=πx2+x22+C.\begin{aligned} \sin^{-1}(\cos x) &= \frac{\pi}{2} + x, \\ \int\sin^{-1}(\cos x)\,dx &= \frac{\pi x}{2} + \frac{x^2}{2} + C. \end{aligned}

    Together, on [−π,π][-\pi, \pi]⁠, sin⁡−1(cos⁡x)=π2−∣x∣\sin^{-1}(\cos x) = \frac{\pi}{2} - |x|⁠, and the function repeats with period 2π2\pi because cos⁡x\cos x does: its graph is the zigzag below.

    −π −π/2 π/2 π 3π/2 2π π/2 −π/2 x y y = π/2 − x on [0, π] y = sin⁻¹(cos x) line continued line continued
    y=sin⁡−1(cos⁡x)y = \sin^{-1}(\cos x) is a zigzag between −π2-\frac{\pi}{2} and π2\frac{\pi}{2} with period 2π2\pi⁠. On [0,π][0, \pi] (blue) it is the line y=π2−xy = \frac{\pi}{2} - x⁠; the dashed gold extensions of that line leave the graph at once.

    A single antiderivative valid on all of [−π,π][-\pi, \pi] is πx2−x∣x∣2\frac{\pi x}{2} - \frac{x|x|}{2}⁠, since ddx(x∣x∣)=2∣x∣\frac{d}{dx}\left(x|x|\right) = 2|x|⁠.

    The textbook's answer is the one for 0≤x≤π0 \le x \le \pi⁠, the interval on which π2−x\frac{\pi}{2} - x is the principal value. Saying where an answer holds is part of the answer whenever an inverse trigonometric function is simplified.
    sin⁡(π2+x)=cos⁡x\sin\left(\frac{\pi}{2} + x\right) = \cos x⁠; Section 2.3 again, now with θ=π2+x\theta = \frac{\pi}{2} + x⁠.
Result ∫sin⁡−1(cos⁡x) dx=πx2−x22+C(0≤x≤π).\int\sin^{-1}(\cos x)\,dx = \frac{\pi x}{2} - \frac{x^2}{2} + C \qquad (0 \le x \le \pi).

On [−π,0][-\pi, 0] it is πx2+x22+C\frac{\pi x}{2} + \frac{x^2}{2} + C instead. (Textbook answer: πx2−x22+C\frac{\pi x}{2} - \frac{x^2}{2} + C⁠, which holds for 0≤x≤π0 \le x \le \pi⁠; the book does not state the interval.)

Check it with numbers

Let F(x)=πx2−x22F(x) = \frac{\pi x}{2} - \frac{x^2}{2} and look near x=1x = 1 (inside [0,π][0, \pi]⁠) with h=0.0001h = 0.0001⁠:

  • F(1.0001)≈1.070853401428F(1.0001) \approx 1.070853401428 and F(0.9999)≈1.070739242162F(0.9999) \approx 1.070739242162⁠.
  • Slope estimate: F(1.0001)−F(0.9999)0.0002≈0.570796\dfrac{F(1.0001) - F(0.9999)}{0.0002} \approx 0.570796⁠.
  • Integrand at x=1x = 1⁠: sin⁡−1(cos⁡1)≈0.570796\sin^{-1}(\cos 1) \approx 0.570796⁠.

They agree. At x=−1x = -1⁠, however, F′(−1)=π2+1≈2.570796F'(-1) = \frac{\pi}{2} + 1 \approx 2.570796 while the integrand is 0.5707960.570796⁠: outside [0,π][0, \pi] the textbook form fails, as Step 6 explains.

Common mistakes
  1. Writing sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta) = \theta for every θ\theta⁠. It holds only for θ\theta in [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]⁠; here that means 0≤x≤π0 \le x \le \pi⁠.
  2. Treating sin⁡−1\sin^{-1} as 1sin⁡\frac{1}{\sin}⁠. sin⁡−1(cos⁡x)\sin^{-1}(\cos x) is the inverse sine of cos⁡x\cos x⁠, not 1sin⁡(cos⁡x)\frac{1}{\sin(\cos x)}⁠, and not cot⁡x\cot x⁠.
  3. Giving the answer without its interval. The integrand is a zigzag; one quadratic cannot be its antiderivative everywhere.
What to take away
  1. Simplify an inverse trigonometric expression by making the inside match the outer function, then check the principal range to find where the simplification is valid.
  2. sin⁡−1(cos⁡x)=π2−x\sin^{-1}(\cos x) = \frac{\pi}{2} - x for 0≤x≤π0 \le x \le \pi⁠; more generally it equals π2−∣x∣\frac{\pi}{2} - |x| on [−π,π][-\pi, \pi]⁠.

Every fact used, and where it comes from

FactUsed inSource
Domain [−1,1][-1, 1] and principal range [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] of sin⁡−1\sin^{-1}Steps 1, 3Chapter 2, Section 2.2
sin⁡(π2∓x)=cos⁡x\sin\left(\frac{\pi}{2} \mp x\right) = \cos xSteps 2, 6Class XI trigonometry (Basics: trigonometric identities)
sin⁡−1(sin⁡x)=x\sin^{-1}(\sin x) = x for x∈[−π2,π2]x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]Steps 3, 6Chapter 2, Section 2.3
∫xn dx=xn+1n+1+C\int x^n\,dx = \frac{x^{n+1}}{n+1} + C⁠; ∫dx=x+C\int dx = x + CSteps 4, 6Standard formula (i), Section 7.2
∫\int of a linear combinationStep 4Property (V), Section 7.2.1
Differentiation undoes integrationStep 5Property (I), Section 7.2.1
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 →