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Chapter 7 · Integrals

Miscellaneous · Q30: ∫0π/2sin⁡2x tan⁡−1(sin⁡x) dx\int_0^{\pi/2}\sin 2x\,\tan^{-1}(\sin x)\,dx

Miscellaneous ExerciseSubstitution, then integration by parts · Miscellaneous Exercise
The question

Evaluate the definite integral

∫0π/2sin⁡2x tan⁡−1(sin⁡x) dx\int_0^{\pi/2} \sin 2x\,\tan^{-1}(\sin x)\,dx
Exam answer

Let I=∫0π/2sin⁡2x tan⁡−1(sin⁡x) dxI = \displaystyle\int_0^{\pi/2}\sin 2x\,\tan^{-1}(\sin x)\,dx⁠.

Using sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x⁠, put sin⁡x=t\sin x = t⁠. Then cos⁡x dx=dt\cos x\,dx = dt⁠.

When x=0x = 0⁠, t=0t = 0⁠; when x=π2x = \frac{\pi}{2}⁠, t=1t = 1⁠.

I=∫012ttan⁡−1t dtI = \int_0^1 2t\tan^{-1}t\,dt

Taking tan⁡−1t\tan^{-1}t as the first function and 2t2t as the second, by parts:

I=[t2tan⁡−1t]01−∫01t21+t2 dt=π4−∫01(1−11+t2)dt=π4−[t−tan⁡−1t]01=π4−(1−π4)\begin{aligned} I &= \Big[t^2\tan^{-1}t\Big]_0^1 \\ &\quad - \int_0^1\frac{t^2}{1 + t^2}\,dt \\ &= \frac{\pi}{4} - \int_0^1\left(1 - \frac{1}{1 + t^2}\right)dt \\ &= \frac{\pi}{4} - \Big[t - \tan^{-1}t\Big]_0^1 \\ &= \frac{\pi}{4} - \left(1 - \frac{\pi}{4}\right) \end{aligned}

∴ I=π2−1\therefore\ I = \dfrac{\pi}{2} - 1⁠.

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

The one idea this problem is built on

Two methods, one after the other. First, sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x contains cos⁡x dx\cos x\,dx⁠, the differential of sin⁡x\sin x⁠, so t=sin⁡xt = \sin x turns the integral into ∫2ttan⁡−1t dt\int 2t\tan^{-1}t\,dt⁠. Second, that is a product of an algebraic function and an inverse trigonometric function, which is the classic case for integration by parts with tan⁡−1t\tan^{-1} t as the first function.

Given, to find, and the plan
Given
I=∫0π/2sin⁡2x tan⁡−1(sin⁡x) dxI = \displaystyle\int_0^{\pi/2} \sin 2x\,\tan^{-1}(\sin x)\,dx⁠.
To find
The exact value of II⁠, and a decimal.
Plan
(1) Continuity. (2) Write sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x and substitute t=sin⁡xt = \sin x⁠, changing the limits. (3) Integrate 2ttan⁡−1t2t\tan^{-1}t by parts. (4) Handle the leftover t21+t2\frac{t^2}{1+t^2} by division. (5) Check by differentiating. (6) Evaluate.

Basics you need

1Double angle

sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x⁠. Basics: trigonometric identities

2Substitution with changed limits (Section 7.9)

With t=sin⁡xt = \sin x⁠, dt=cos⁡x dxdt = \cos x\,dx⁠; the limits become sin⁡0=0\sin 0 = 0 and sin⁡π2=1\sin\frac{\pi}{2} = 1⁠. Basics: substitution

3Integration by parts (Section 7.6)

∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du⁠. Choose as uu (the first function) the factor that becomes simpler when differentiated; Remark (iii) of Section 7.6 says to take an inverse trigonometric or logarithmic function first. Lesson: integration by parts

4Two standard facts

ddttan⁡−1t=11+t2\frac{d}{dt}\tan^{-1}t = \frac{1}{1+t^2}⁠, so ∫dt1+t2=tan⁡−1t+C\int\frac{dt}{1+t^2} = \tan^{-1}t + C (standard formula (x), Section 7.2); and tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠. Basics: integration undoes differentiation

5Dividing out a fraction

t21+t2\frac{t^2}{1+t^2} =(1+t2)−11+t2= \frac{(1+t^2) - 1}{1+t^2} =1−11+t2= 1 - \frac{1}{1+t^2}⁠. Basics: splitting fractions

Solution

  1. The integrand is continuous on [0,π2][0, \frac{\pi}{2}]

    sin⁡2x\sin 2x and sin⁡x\sin x are continuous everywhere, and tan⁡−1\tan^{-1} is continuous on all of R\R⁠. So tan⁡−1(sin⁡x)\tan^{-1}(\sin x)⁠, a composition of continuous functions, is continuous, and so is the product sin⁡2x tan⁡−1(sin⁡x)\sin 2x\,\tan^{-1}(\sin x)⁠.

    Theorem 2 of Section 7.8.3 needs continuity on the closed interval. There are no denominators, square roots or logarithms here, so nothing can go wrong.
    Sums, products and compositions of continuous functions are continuous (Chapter 5, Section 5.2). Lesson: definite integrals

  2. Substitute t=sin⁡xt = \sin x

    Write sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x⁠:

    I=∫0π/22sin⁡x tan⁡−1(sin⁡x) cos⁡x dx.I = \int_0^{\pi/2} 2\sin x\,\tan^{-1}(\sin x)\,\cos x\,dx .

    Put t=sin⁡xt = \sin x⁠, dt=cos⁡x dxdt = \cos x\,dx⁠. At x=0x = 0⁠, t=0t = 0⁠; at x=π2x = \frac{\pi}{2}⁠, t=1t = 1⁠. So

    I=∫012t tan⁡−1t dt.I = \int_0^1 2t\,\tan^{-1}t\,dt .
    The integrand contains sin⁡x\sin x inside tan⁡−1\tan^{-1}⁠, and once sin⁡2x\sin 2x is opened up, the factor cos⁡x\cos x is available to form dtdt⁠. Everything else is a function of sin⁡x\sin x⁠. That is exactly the pattern substitution is built for. The substitution is valid on the whole interval: sin⁡x\sin x is differentiable and increasing on [0,π2][0, \frac{\pi}{2}] (its derivative cos⁡x≥0\cos x \ge 0⁠), so it maps [0,π2][0, \frac{\pi}{2}] onto [0,1][0, 1] once.
    sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x⁠; Section 7.9. Lesson: integration by substitution

  3. Integration by parts with u=tan⁡−1tu = \tan^{-1} t

    Choose

    u=tan⁡−1t,du=dt1+t2,dv=2t dt,v=t2.\begin{aligned} u &= \tan^{-1}t, & du &= \frac{dt}{1+t^2}, \\ dv &= 2t\,dt, & v &= t^2 . \end{aligned}

    Then ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du gives

    ∫2ttan⁡−1t dt=t2tan⁡−1t−∫t21+t2 dt.\begin{aligned} &\int 2t\tan^{-1}t\,dt \\ &= t^2\tan^{-1}t - \int\frac{t^2}{1+t^2}\,dt . \end{aligned}
    Why tan⁡−1t\tan^{-1}t first: we know how to differentiate it (into a simple fraction) but cannot integrate it directly, while 2t2t is easy to integrate. This is Remark (iii) of Section 7.6 (inverse trigonometric functions are taken as the first function; the "ILATE" order Inverse, Logarithmic, Algebraic, Trigonometric, Exponential is a memory aid for the same advice). The other choice, u=2tu = 2t⁠, would require ∫tan⁡−1t dt\int\tan^{-1}t\,dt⁠, a harder integral.
    Section 7.6, formula (1): ∫udvdxdx=uv−∫vdudxdx\int u\frac{dv}{dx}dx = uv - \int v\frac{du}{dx}dx⁠. By Remark (ii) there, no constant is needed in vv⁠. Lesson: integration by parts

  4. The leftover integral: divide first

    ∫t21+t2 dt=∫(1−11+t2)dt=t−tan⁡−1t.\begin{aligned} \int\frac{t^2}{1+t^2}\,dt &= \int\left(1 - \frac{1}{1+t^2}\right)dt \\ &= t - \tan^{-1}t . \end{aligned}

    Putting this into Step 3:

    ∫2ttan⁡−1t dt=t2tan⁡−1t−t+tan⁡−1t=(t2+1)tan⁡−1t−t+C.\begin{aligned} &\int 2t\tan^{-1}t\,dt \\ &= t^2\tan^{-1}t - t + \tan^{-1}t \\ &= (t^2 + 1)\tan^{-1}t - t + C . \end{aligned}
    t21+t2\frac{t^2}{1+t^2} has equal degrees on top and bottom, so we divide (add and subtract 1 in the numerator) to get a polynomial plus a proper fraction. The proper fraction 11+t2\frac{1}{1+t^2} is standard formula (x).
    Basics: splitting fractions

  5. Check: differentiate the antiderivative

    Let H(t)=(t2+1)tan⁡−1t−tH(t) = (t^2 + 1)\tan^{-1}t - t⁠. By the product rule:

    H′(t)=2ttan⁡−1t+(t2+1)⋅11+t2−1=2ttan⁡−1t+1−1=2ttan⁡−1t ✓\begin{aligned} H'(t) &= 2t\tan^{-1}t + (t^2 + 1)\cdot\frac{1}{1+t^2} - 1 \\ &= 2t\tan^{-1}t + 1 - 1 = 2t\tan^{-1}t \ ✓ \end{aligned}

    valid for every real tt⁠. In terms of xx⁠, H(sin⁡x)=(sin⁡2x+1)tan⁡−1(sin⁡x)−sin⁡xH(\sin x) = (\sin^2 x + 1)\tan^{-1}(\sin x) - \sin x⁠, and the chain rule gives 2sin⁡xtan⁡−1(sin⁡x)⋅cos⁡x=sin⁡2xtan⁡−1(sin⁡x)2\sin x\tan^{-1}(\sin x)\cdot\cos x = \sin 2x\tan^{-1}(\sin x)⁠, the original integrand.

    The +1+1 from differentiating tan⁡−1t\tan^{-1}t (times t2+1t^2 + 1⁠) cancels the −1-1 from differentiating −t-t⁠. That cancellation is a good sign that both pieces of Step 4 are right.
    Basics: integration undoes differentiation

  6. Evaluate between 0 and 1

    H(1)=2tan⁡−11−1=2⋅π4−1=π2−1H(1) = 2\tan^{-1}1 - 1 = 2\cdot\frac{\pi}{4} - 1 = \frac{\pi}{2} - 1⁠.

    H(0)=1⋅tan⁡−10−0=0H(0) = 1\cdot\tan^{-1}0 - 0 = 0⁠.

    I=H(1)−H(0)=π2−1≈0.570796.I = H(1) - H(0) = \frac{\pi}{2} - 1 \approx 0.570796 .
    Theorem 2, Section 7.8.3. The answer is positive, as it must be: on (0,π2)(0, \frac{\pi}{2}) both sin⁡2x\sin 2x and tan⁡−1(sin⁡x)\tan^{-1}(\sin x) are positive. Size check: 2ttan⁡−1t2t\tan^{-1}t lies between 0 and 2⋅π4≈1.5712\cdot\frac{\pi}{4} \approx 1.571 on [0,1][0, 1] and is increasing, so an answer near 0.57 is reasonable.
    tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4} (principal value). Lesson: definite integrals
Result

∫0π/2sin⁡2x tan⁡−1(sin⁡x) dx=π2−1≈0.570796\begin{aligned} &\int_0^{\pi/2} \sin 2x\,\tan^{-1}(\sin x)\,dx \\ &= \frac{\pi}{2} - 1 \approx 0.570796 \end{aligned} (Textbook answer: π2−1\frac{\pi}{2} - 1⁠.)

Check it with numbers

Numerical integration (adaptive quadrature) of the original integrand over [0,π2][0, \frac{\pi}{2}] gives 0.5707963268, and π2−1\frac{\pi}{2} - 1 ≈1.5707963268−1\approx 1.5707963268 - 1 =0.5707963268= 0.5707963268 ✓. The integral after substitution, ∫012ttan⁡−1t dt\int_0^1 2t\tan^{-1}t\,dt⁠, gives the same 0.5707963268.

(Q37 of this exercise, ∫01sin⁡−1x dx\int_0^1\sin^{-1}x\,dx⁠, happens to have the same value π2−1\frac{\pi}{2} - 1⁠.)

Common mistakes
  1. Choosing u=2tu = 2t in the parts step. Then you need ∫tan⁡−1t dt\int\tan^{-1}t\,dt⁠, which itself needs parts.
  2. Writing ∫t21+t2dt=tan⁡−1t\int\frac{t^2}{1+t^2}dt = \tan^{-1}t⁠. The fraction must be divided out first: it equals 1−11+t21 - \frac{1}{1+t^2}⁠.
  3. Keeping the xx-limits 00 and π2\frac{\pi}{2} after switching to tt⁠. The tt-limits are 00 and 11⁠.
  4. Losing the factor 2 from sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x⁠: that gives half the answer.
What to take away
  1. Look for the derivative of an inner function first (here cos⁡x\cos x for sin⁡x\sin x⁠); substitution often turns a strange integral into a familiar one.
  2. For ∫(polynomial)⋅tan⁡−1t dt\int(\text{polynomial})\cdot\tan^{-1}t\,dt⁠, take tan⁡−1t\tan^{-1}t as the first function, then divide out the rational leftover.
  3. ∫2ttan⁡−1t dt=(t2+1)tan⁡−1t−t+C\int 2t\tan^{-1}t\,dt = (t^2 + 1)\tan^{-1}t - t + C is worth remembering.

Every fact used, and where it comes from

FactUsed inSource
Continuity of sums, products, compositionsStep 1Chapter 5, Section 5.2
sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos xStep 2Trigonometric functions (Class XI)
Substitution with changed limitsStep 2Section 7.9 (Lesson: substitution)
Integration by parts; choice of first functionStep 3Section 7.6, formula (1) and Remarks (ii), (iii)
∫dt1+t2=tan⁡−1t+C\int\frac{dt}{1+t^2} = \tan^{-1}t + CStep 4Standard formula (x), Section 7.2
Product rule; ddttan⁡−1t=11+t2\frac{d}{dt}\tan^{-1}t = \frac{1}{1+t^2}Step 5Chapter 5, Sections 5.3 and 5.3.3
F(b)−F(a)F(b) - F(a)⁠; tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}Step 6Theorem 2, Section 7.8.3; Chapter 2
All decimalsStep 6, CheckRecomputed (mpmath quadrature, 30 digits) in the site's verification script
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