Chapter 7 · Integrals

Exercise 7.3 · Q18: $\int\frac{\cos 2x+2\sin^2x}{\cos^2x}\,dx$

Exercise 7.3Integration using trigonometric identities · Section 7.3.2

The question

Find the integral of the function

\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}.

Exam answer

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

Using $\cos 2x = 1 - 2\sin^2 x$ ⁠,

\begin{aligned} I &= \int\frac{1 - 2\sin^2 x + 2\sin^2 x}{\cos^2 x}\,dx \\ &= \int\frac{1}{\cos^2 x}\,dx = \int\sec^2 x\,dx \end{aligned}

$\therefore\ I = \tan x + C$ ⁠.

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

The one idea this problem is built on

One of the forms of the double-angle identity is $\cos 2x = 1 - 2\sin^2 x$ ⁠. Then the numerator is $1 - 2\sin^2 x + 2\sin^2 x = 1$ ⁠, the whole integrand collapses to $\frac{1}{\cos^2 x} = \sec^2 x$ ⁠, and $\int\sec^2 x\,dx = \tan x + C$ ⁠.

Given, to find, and the plan

Given: $f(x) = \dfrac{\cos 2x + 2\sin^2 x}{\cos^2 x}$ ⁠, defined wherever $\cos x \ne 0$ ⁠.
To find: $\displaystyle\int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x}\,dx$ ⁠.
Plan: (1) Find the domain. (2) Simplify the numerator with $\cos 2x = 1 - 2\sin^2 x$ ⁠. (3) Integrate $\sec^2 x$ ⁠. (4) Differentiate back.

Basics you need

1The three forms of $\cos 2x$

$\cos 2x = \cos^2 x - \sin^2 x$ ⁠, which also equals $2\cos^2 x - 1$ and $1 - 2\sin^2 x$ ⁠. Pick the form that matches the rest of the expression; here the numerator already has $2\sin^2 x$ ⁠. Basics: trigonometric identities for integrals

2 $\int\sec^2 x\,dx = \tan x + C$

Standard formula (iv), Section 7.2, because $\frac{d}{dx}\tan x = \sec^2 x$ ⁠. Basics: integration undoes differentiation

Solution

Where is the integrand defined?

The denominator is $\cos^2 x$ ⁠, so we need $\cos x \ne 0$ ⁠:
$x \ne \frac{\pi}{2} + n\pi, \quad n \text{ an integer}.$
The numerator is defined for every $x$ ⁠; only the denominator can fail. The answer will be valid on each interval between consecutive such points, for example $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ ⁠.
Simplify the numerator

Use $\cos 2x = 1 - 2\sin^2 x$ ⁠:
$\begin{aligned} \cos 2x + 2\sin^2 x &= (1 - 2\sin^2 x) + 2\sin^2 x \\ &= 1. \end{aligned}$
Therefore
$\frac{\cos 2x + 2\sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x.$
The numerator contains $2\sin^2 x$ ⁠, so the form of $\cos 2x$ written with $\sin^2 x$ is the natural choice: the two $\sin^2 x$ terms cancel exactly. (Using $\cos^2 x - \sin^2 x$ also works: the numerator becomes $\cos^2 x + \sin^2 x = 1$ ⁠.)

$\cos 2x = 1 - 2\sin^2 x$ (Class XI; compare Example 7(i), Section 7.3.2); $\sec x = \frac{1}{\cos x}$ ⁠. Basics: trigonometric identities for integrals
Integrate
$\begin{aligned} \int \frac{\cos 2x + 2\sin^2 x}{\cos^2 x}\,dx &= \int\sec^2 x\,dx \\ &= \tan x + C. \end{aligned}$
After the simplification the integrand is a standard derivative: $\tan x$ differentiates to $\sec^2 x$ ⁠, so $\tan x$ is an antiderivative, and adding $C$ gives them all.

Standard formula (iv), Section 7.2. Lesson: integration as anti-differentiation
Check: differentiate the answer back
$\begin{aligned} \frac{d}{dx}\tan x &= \sec^2 x = \frac{1}{\cos^2 x} \\ &= \frac{(1 - 2\sin^2 x) + 2\sin^2 x}{\cos^2 x} \\ &= \frac{\cos 2x + 2\sin^2 x}{\cos^2 x}. \end{aligned}$
The middle line writes $1$ as $(1 - 2\sin^2 x) + 2\sin^2 x$ ⁠, and the last uses $1 - 2\sin^2 x = \cos 2x$ ⁠. This is the integrand, so the answer is correct on every interval where $\cos x \ne 0$ ⁠.

List in Section 7.2, (iv); Property (I), Section 7.2.1. Basics: integration undoes differentiation

Result

\int\frac{\cos 2x + 2\sin^2 x}{\cos^2 x}\,dx = \tan x + C,

valid on any interval on which $\cos x \ne 0$ ⁠. (Textbook answer: $\tan x + C$ ⁠.)

Check it with numbers

Let $F(x) = \tan x$ and look near $x = 0.4$ with $h = 0.0001$ ⁠:

$F(0.4001) \approx 0.422911099133$ and $F(0.3999) \approx 0.422675348311$ ⁠.
Slope estimate: $\dfrac{F(0.4001) - F(0.3999)}{0.0002} \approx 1.178754$ ⁠.
Integrand at $x = 0.4$ ⁠: $\dfrac{\cos 0.8 + 2\sin^2 0.4}{\cos^2 0.4} \approx 1.178754$ ⁠, the same as $\sec^2 0.4$ ⁠.

They agree.

Common mistakes

Splitting first and integrating $\frac{\cos 2x}{\cos^2 x}$ on its own. It can be done, but it is far longer. Simplify the numerator before integrating.
Using $\cos 2x = 2\sin^2 x - 1$ ⁠. The sign is wrong; the correct form is $1 - 2\sin^2 x$ ⁠.
Writing $\int\sec^2 x\,dx = \sec x$ or $\frac{\sec^3 x}{3}$ ⁠. The power rule does not apply to $\sec^2 x$ ⁠; the standard formula gives $\tan x$ ⁠.

What to take away

Always look for simplification before integrating: here the whole numerator is the constant 1.
Choose the form of $\cos 2x$ that cancels with the terms already present.

Every fact used, and where it comes from

Fact	Used in	Source
$\cos 2x = 1 - 2\sin^2 x$	Steps 2, 4	Class XI trigonometry; compare Example 7(i), Section 7.3.2 (Basics: trigonometric identities)
$\sec x = \frac{1}{\cos x}$	Step 2	Definition
$\int\sec^2 x\,dx = \tan x + C$	Step 3	Standard formula (iv), Section 7.2
Differentiation undoes integration	Step 4	Property (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 →

Exercise 7.3 · Q18: ∫cos⁡2x+2sin⁡2xcos⁡2x dx\int\frac{\cos 2x+2\sin^2x}{\cos^2x}\,dx∫cos2xcos2x+2sin2x​dx

Basics you need

Solution

Where is the integrand defined?

Simplify the numerator

Integrate

Check: differentiate the answer back

Every fact used, and where it comes from

Exercise 7.3 · Q18: $\int\frac{\cos 2x+2\sin^2x}{\cos^2x}\,dx$