Chapter 7 · Integrals

Exercise 7.9 · Q5: $\int_0^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx$

Exercise 7.9Definite integrals by substitution · Section 7.9

The question

Evaluate the integral using substitution:

\int_0^{\frac{\pi}{2}} \frac{\sin x}{1 + \cos^2 x}\,dx.

Exam answer

Let $I = \displaystyle\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}\,dx$ ⁠.

Put $\cos x = t$ ⁠. Then $-\sin x\,dx = dt$ ⁠.

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

\therefore\ I = \int_1^0\frac{-dt}{1 + t^2}

Swapping the limits removes the minus sign:

\begin{aligned} I &= \int_0^1\frac{dt}{1 + t^2} = \Big[\tan^{-1}t\Big]_0^1 \\ &= \tan^{-1}1 - \tan^{-1}0 \end{aligned}

$\therefore\ I = \dfrac{\pi}{4}$ ⁠.

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

The one idea this problem is built on

$\sin x\,dx$ is (minus) the differential of $\cos x$ ⁠. With $t = \cos x$ the integral becomes $-\int\frac{dt}{1 + t^2}$ ⁠, an inverse tangent. Because cosine decreases, the new limits come out "upside down" (from 1 to 0), and the minus sign turns them the right way round.

Given, to find, and the plan

Given: The integrand $\frac{\sin x}{1 + \cos^2 x}$ on $\left[0, \frac{\pi}{2}\right]$ ⁠.
To find: The exact value of the integral, and a decimal.
Plan: (1) Check continuity. (2) Substitute $t = \cos x$ ⁠, write $dt$ ⁠, change the limits. (3) Integrate $\frac{1}{1+t^2}$ and evaluate, keeping track of the minus sign. (4) Check with an antiderivative in $x$ ⁠, and numerically.

Basics you need

1Derivative of $\cos x$

$\frac{d}{dx}\cos x = -\sin x$ ⁠, so with $t = \cos x$ ⁠, $dt = -\sin x\,dx$ ⁠. Basics: substitution (the chain rule backwards)

2Inverse tangent integral

$\int\frac{dt}{1 + t^2} = \tan^{-1}t + C$ (standard formula (x), Section 7.2). Lesson: integrals of particular functions

3Substitution with new limits

$\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt$ ⁠, even when $g(b) < g(a)$ (Section 7.9). Lesson: integration by substitution

4The second fundamental theorem

$\int_a^b f = F(b) - F(a)$ when $f$ is continuous on $[a, b]$ and $F' = f$ (Theorem 2, Section 7.8.3). Lesson: definite integrals

Solution

Check the integrand is continuous on $\left[0, \frac{\pi}{2}\right]$

$\sin x$ and $\cos^2 x$ are continuous for every $x$ ⁠, and $1 + \cos^2 x \ge 1$ is never 0. So the quotient is continuous on $\left[0, \frac{\pi}{2}\right]$ ⁠.

The fundamental theorem needs a continuous integrand on the closed interval. With a denominator of at least 1, nothing can go wrong.

A quotient of continuous functions is continuous where the denominator is not 0 (Chapter 5, Section 5.2.1). Lesson: the fundamental theorem
Substitute $t = \cos x$ and change the limits

Then $dt = -\sin x\,dx$ ⁠, so
$\sin x\,dx = -\,dt, \qquad 1 + \cos^2 x = 1 + t^2.$
New limits. When $x = 0$ ⁠, $t = \cos 0 = 1$ ⁠. When $x = \frac{\pi}{2}$ ⁠, $t = \cos\frac{\pi}{2} = 0$ ⁠.
$\int_0^{\frac{\pi}{2}}\frac{\sin x\,dx}{1 + \cos^2 x} = \int_1^0\frac{-\,dt}{1 + t^2}.$
Why it is valid on the whole interval. $t = \cos x$ has the continuous derivative $-\sin x$ ⁠, which is $\le 0$ on $\left[0, \frac{\pi}{2}\right]$ ⁠. So $t$ decreases steadily from 1 to 0, and $\frac{1}{1+t^2}$ is continuous for every $t$ ⁠.

The numerator $\sin x$ is, apart from its sign, the derivative of $\cos x$ ⁠, and the denominator is built from $\cos x$ ⁠. The lower limit 0 maps to $t = 1$ and the upper limit $\frac{\pi}{2}$ to $t = 0$ ⁠: the new limits keep their positions (lower first), even though 1 is bigger than 0.

$\cos 0 = 1$ ⁠, $\cos\frac{\pi}{2} = 0$ ⁠. Basics: trigonometry
Integrate in $t$
$\begin{aligned}\int_1^0\frac{-\,dt}{1 + t^2} &= \Big[-\tan^{-1}t\Big]_1^0 \\ &= \left(-\tan^{-1}0\right) - \left(-\tan^{-1}1\right) \\ &= 0 + \frac{\pi}{4} = \frac{\pi}{4}.\end{aligned}$
In decimals, $\frac{\pi}{4} \approx 0.7853982$ ⁠.

We evaluate at the upper limit (0) minus the lower limit (1), exactly as written. Equivalently, swapping the limits removes the minus sign: $\int_1^0 -g = \int_0^1 g$ (property P1 of Section 7.10).

$\tan^{-1}0 = 0$ and $\tan^{-1}1 = \frac{\pi}{4}$ (principal values, Chapter 2). Lesson: integrals of particular functions
Same result by staying in $x$

Returning to $x$ ⁠: $-\tan^{-1}t$ becomes
$F(x) = -\tan^{-1}(\cos x).$
Check: differentiate. Chain rule with outer $-\tan^{-1}u$ and inner $u = \cos x$ ⁠:
$F'(x) = -\frac{1}{1 + \cos^2 x}\cdot(-\sin x) = \frac{\sin x}{1 + \cos^2 x},$
the integrand. Then $F\left(\frac{\pi}{2}\right) - F(0) = -\tan^{-1}0 + \tan^{-1}1 = \frac{\pi}{4}$ ⁠.

The two minus signs (from $-\tan^{-1}$ and from $\frac{d}{dx}\cos x = -\sin x$ ⁠) cancel, which is exactly why the substitution produced a minus sign that later disappeared.

$\frac{d}{du}\tan^{-1}u = \frac{1}{1+u^2}$ (Chapter 5, Section 5.3.3). Basics: the chain rule backwards
The region

$y = \frac{\sin x}{1+\cos^2x}$ rises from 0 at $x = 0$ to 1 at $x = \frac{\pi}{2}$ ⁠. The shaded area is $\frac{\pi}{4} \approx 0.785$ ⁠.

The integrand is $\ge 0$ on the interval, so the answer is an area and must be positive, which our careful handling of the minus sign delivers. The region lies inside the $\frac{\pi}{2}$ by 1 rectangle (area 1.571), and about half of it is filled.

For a function that is never negative, the definite integral is the area under its graph. Basics: area as a sum of thin strips

Result

$\int_0^{\frac{\pi}{2}}\frac{\sin x}{1 + \cos^2 x}\,dx = \frac{\pi}{4} \approx 0.7853982.$ (Textbook answer: $\frac{\pi}{4}$ ⁠.)

Check it with numbers

A numerical integration of $\frac{\sin x}{1+\cos^2 x}$ over $\left[0, \frac{\pi}{2}\right]$ gives 0.7853982, and $\pi \div 4 \approx 0.7853982$ ⁠. The area is exactly half of the rectangle $\frac{\pi}{2}\times 1$ ⁠: $\frac{\pi/4}{\pi/2} = 0.5$ ⁠.

Common mistakes

Losing the minus sign. $dt = -\sin x\,dx$ ⁠. Dropping the minus and keeping the limits 1 to 0 gives $-\frac{\pi}{4}$ ⁠, a negative "area" for a positive integrand.
Swapping the limits and also keeping the minus. Do one or the other: $\int_1^0(-g) = \int_0^1 g$ ⁠.
Writing $\tan^{-1}1 = 1$ ⁠. $\tan^{-1}1 = \frac{\pi}{4}$ because $\tan\frac{\pi}{4} = 1$ ⁠.

What to take away

$\sin x\,dx$ next to a function of $\cos x$ signals $t = \cos x$ (and $\cos x\,dx$ next to a function of $\sin x$ signals $t = \sin x$ ⁠).
A decreasing substitution gives "upside-down" limits; keep them, and let the minus sign from $dt$ fix the order.
A positive integrand must give a positive answer: a quick sign check.

Every fact used, and where it comes from

Fact	Used in	Source
Continuity of quotients	Step 1	Chapter 5, Section 5.2.1
$\frac{d}{dx}\cos x = -\sin x$	Steps 2 and 4	Chapter 5; standard formula (iii), Section 7.2
Substitution with change of limits	Step 2	Section 7.9 and its Note (Lesson)
$\int\frac{dt}{1+t^2} = \tan^{-1}t + C$	Step 3	Standard formula (x), Section 7.2 (Lesson)
$\tan^{-1}0 = 0$ ⁠, $\tan^{-1}1 = \frac{\pi}{4}$	Steps 3 and 4	Chapter 2, principal values
$\int_b^a f = -\int_a^b f$	Step 3 (remark)	Property P1, Section 7.10 (Lesson)
$\int_a^b f = F(b) - F(a)$	Steps 3 and 4	Theorem 2, Section 7.8.3 (Lesson)

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Exercise 7.9 · Q5: ∫0π/2sin⁡x1+cos⁡2x dx\int_0^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx∫0π/2​1+cos2xsinx​dx

Basics you need

Solution

Check the integrand is continuous on [0,π2]\left[0, \frac{\pi}{2}\right][0,2π​]

Substitute t=cos⁡xt = \cos xt=cosx and change the limits

Integrate in ttt

Same result by staying in xxx

The region

Every fact used, and where it comes from

Exercise 7.9 · Q5: $\int_0^{\pi/2}\frac{\sin x}{1+\cos^2x}\,dx$

Check the integrand is continuous on $\left[0, \frac{\pi}{2}\right]$

Substitute $t = \cos x$ and change the limits

Integrate in $t$

Same result by staying in $x$