Chapter 7 · Integrals

Exercise 7.10 · Q15: $\int_0^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx$

Exercise 7.10Properties of definite integrals (P4) · Section 7.10

The question

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

15. $\displaystyle\int_0^{\pi/2}\frac{\sin x - \cos x}{1 + \sin x\cos x}\,dx$

Exam answer

Let

I = \int_0^{\pi/2}\frac{\sin x - \cos x}{1 + \sin x\cos x}\,dx\;\ldots(\text{i})

Using $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ ⁠, and $\sin\left(\frac{\pi}{2} - x\right) = \cos x$ ⁠, $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ ⁠,

I = \int_0^{\pi/2}\frac{\cos x - \sin x}{1 + \cos x\sin x}\,dx\;\ldots(\text{ii})

Adding (i) and (ii),

2I = \int_0^{\pi/2} 0\,dx = 0

$\therefore\ I = 0$ ⁠.

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

The one idea this problem is built on

On $[0, \frac{\pi}{2}]$ ⁠, P4 replaces $x$ by $\frac{\pi}{2} - x$ ⁠, which swaps $\sin x$ and $\cos x$ ⁠. The denominator $1 + \sin x\cos x$ does not notice the swap, but the numerator $\sin x - \cos x$ changes sign. So P4 says $I = -I$ ⁠, and the only number equal to its own negative is 0.

Given, to find, and the plan

Given: $f(x) = \dfrac{\sin x - \cos x}{1 + \sin x\cos x}$ on $[0, \frac{\pi}{2}]$ ⁠.
To find: $I = \displaystyle\int_0^{\pi/2} f(x)\,dx$ ⁠.
Plan: (1) Continuity. (2) P4 with $a = \frac{\pi}{2}$ and the relation $f(\frac{\pi}{2} - x) = -f(x)$ ⁠. (3) Solve $I = -I$ ⁠. (4) Confirm with an antiderivative.

Basics you need

1Property P4 (Section 7.10)

$\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ for $f$ continuous on $[0, a]$ ⁠. Lesson: properties of definite integrals

2Complementary angles

$\sin\left(\frac{\pi}{2} - x\right) = \cos x$ and $\cos\left(\frac{\pi}{2} - x\right) = \sin x$ ⁠; also $\sin 2x = 2\sin x\cos x$ ⁠. Basics: trigonometric identities

3Point symmetry

If $f(a - x) = -f(x)$ ⁠, the graph on $[0, a]$ is unchanged by a half turn about $(\frac{a}{2}, 0)$ ⁠, so the areas above and below the axis are equal. Basics: symmetry and mirror images

4Derivative of $\tan^{-1}$

$\frac{d}{dx}\tan^{-1}u = \frac{1}{1 + u^2}\,\frac{du}{dx}$ (used in the check). Basics: integration undoes differentiation

Solution

The integrand is continuous on $[0, \frac{\pi}{2}]$

Using $\sin x\cos x = \frac12\sin 2x$ ⁠, the denominator is
$1 + \sin x\cos x = 1 + \tfrac12\sin 2x .$
On $[0, \frac{\pi}{2}]$ ⁠, $2x$ runs over $[0, \pi]$ ⁠, where $\sin 2x \ge 0$ ⁠. So the denominator is at least 1 and never 0. The integrand, a quotient of continuous functions, is continuous on the closed interval.

P4 (and Theorem 2 in the check) need a function continuous on the closed interval. A zero of the denominator would break this; Step 1 rules it out.

$\sin 2x = 2\sin x\cos x$ ⁠; $\sin\theta \ge 0$ for $0 \le \theta \le \pi$ ⁠. Basics: trigonometric identities
P4 with $a = \frac{\pi}{2}$ ⁠: the integrand changes sign

Replace $x$ by $\frac{\pi}{2} - x$ ⁠. Sine and cosine swap:
$\begin{aligned}f\!\left(\tfrac{\pi}{2} - x\right) &= \frac{\cos x - \sin x}{1 + \cos x\sin x} \\ &= -\frac{\sin x - \cos x}{1 + \sin x\cos x} = -f(x) .\end{aligned}$
P4 therefore gives
$I = \int_0^{\pi/2} f\!\left(\tfrac{\pi}{2} - x\right)dx = \int_0^{\pi/2}\big(-f(x)\big)\,dx = -I .$
P4 holds for every continuous $f$ (Step 1), so no further condition is needed. The relation $f(\frac{\pi}{2} - x) = -f(x)$ is exactly the condition of the second case of P6 (with $2a = \frac{\pi}{2}$ ⁠), so P6 would give 0 at once as well; here we see it directly from P4.

$\sin(\frac{\pi}{2} - x) = \cos x$ ⁠, $\cos(\frac{\pi}{2} - x) = \sin x$ ⁠; a minus sign comes out of an integral as a constant factor $-1$ ⁠.
Solve $I = -I$

Add $I$ to both sides:
$2I = 0, \qquad I = 0 .$

The integrand on $[0, \frac{\pi}{2}]$ ⁠: negative on $[0, \frac{\pi}{4})$ (rose), positive on $(\frac{\pi}{4}, \frac{\pi}{2}]$ (green). A half turn about $(\frac{\pi}{4}, 0)$ maps one region onto the other, so the signed areas cancel.

$I$ is a finite number (the integrand is continuous on a closed interval), so it is legitimate to add $I$ to both sides. The figure shows why the answer is 0: the area below the axis on the left equals the area above it on the right.

Basics: area as a sum of thin strips (signed area)
Check: an antiderivative, differentiated back

Let $u = \sin x + \cos x$ ⁠. Then $u^2 = 1 + 2\sin x\cos x$ ⁠, so $1 + \sin x\cos x = \frac{1 + u^2}{2}$ ⁠, and $\frac{du}{dx} = \cos x - \sin x$ ⁠. This suggests
$G(x) = -2\tan^{-1}(\sin x + \cos x) .$
Differentiate back (chain rule):
$\begin{aligned}G'(x) &= -2\cdot\frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \\ &= \frac{2(\sin x - \cos x)}{2 + 2\sin x\cos x} \\ &= \frac{\sin x - \cos x}{1 + \sin x\cos x} = f(x) .\end{aligned}$
By Theorem 2:
$\begin{aligned}I &= G\!\left(\tfrac{\pi}{2}\right) - G(0) \\ &= -2\tan^{-1}1 + 2\tan^{-1}1 = 0 .\end{aligned}$
$u = \sin x + \cos x$ equals 1 at both ends ( $x = 0$ and $x = \frac{\pi}{2}$ ⁠), so $G$ takes the same value at both ends and the integral is 0. This route does not use P4, so it independently confirms Step 3. $G$ is defined and differentiable on the whole interval, since $\tan^{-1}$ is defined for every real number.

$(\sin x + \cos x)^2$ expands to $\sin^2 x + \cos^2 x$ plus $2\sin x\cos x$ ⁠, that is, $1 + 2\sin x\cos x$ ⁠. Basics: substitution (the chain rule backwards)

Result

$\int_0^{\pi/2}\frac{\sin x - \cos x}{1 + \sin x\cos x}\,dx = 0$ (Textbook answer: 0.)

Check it with numbers

Numerical integration over $[0, \frac{\pi}{2}]$ gives 0 (to more than 25 decimal places). Over the left half $[0, \frac{\pi}{4}]$ alone it gives $-0.339837$ ⁠, which matches $G(\frac{\pi}{4}) - G(0)$ $= \frac{\pi}{2} - 2\tan^{-1}\sqrt2$ $\approx -0.339837$ ⁠; the right half gives $+0.339837$ ⁠.

The relation of Step 2 at $x = 0.3$ ⁠: $f(0.3) \approx -0.514548$ and $f(\frac{\pi}{2} - 0.3) \approx 0.514548$ ⁠.

Common mistakes

Adding the two forms and getting $2I = \int 0\,dx$ without seeing why. It is correct, but the clearer statement is $f(\frac{\pi}{2} - x) = -f(x)$ ⁠, so $I = -I$ ⁠.
Thinking the denominator changes under P4. $\sin x\cos x$ is symmetric in $\sin$ and $\cos$ ⁠, so swapping them leaves it unchanged.
Concluding there is no area. The integral is 0 because equal areas lie above and below the axis.

What to take away

On $[0, \frac{\pi}{2}]$ ⁠, P4 swaps $\sin x$ and $\cos x$ ⁠. If the integrand is antisymmetric under that swap, the integral is 0. Q21 works the same way.
If it is unchanged by the swap, add the two forms instead, as in Q3.

Every fact used, and where it comes from

Fact	Used in	Source
A definite integral needs $f$ continuous on $[a, b]$	Step 1	Remark (iv) after Theorem 2, Section 7.8.3
$\sin 2x = 2\sin x\cos x$ ⁠; complementary angles	Steps 1, 2	Class XI trigonometry
P4 (and P6, second case)	Step 2	Section 7.10, properties P4 and P6 (Lesson: properties)
$\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}$ ⁠; chain rule	Step 4	Chapter 5, Section 5.3.3; Section 5.3.1
Theorem 2 (second fundamental theorem)	Step 4	Section 7.8.3

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Exercise 7.10 · Q15: ∫0π/2sin⁡x−cos⁡x1+sin⁡xcos⁡x dx\int_0^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx∫0π/2​1+sinxcosxsinx−cosx​dx

Basics you need

Solution

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

P4 with a=π2a = \frac{\pi}{2}a=2π​⁠: the integrand changes sign

Solve I=−II = -II=−I

Check: an antiderivative, differentiated back

Every fact used, and where it comes from

Exercise 7.10 · Q15: $\int_0^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx$

The integrand is continuous on $[0, \frac{\pi}{2}]$

P4 with $a = \frac{\pi}{2}$ ⁠: the integrand changes sign

Solve $I = -I$