Antiderivatives of trig, exponential and log functions
The power rule run backward handles every power of x except one, and it says nothing about sine, the exponential or the arctangent. Which functions undo those derivatives, and what fills the gap left by one over x?
Where you are 16 in Integrals
Why this exists
The power rule for antiderivatives finds \int x^n\,dx for every real n except n = -1, where it would divide by zero. Yet \dfrac{1}{x} is a perfectly good function on each side of 0, and its graph has a slope at every point, so something must have it as a derivative. The same lesson has nothing to say about \sin x, e^x or \dfrac{1}{1 + x^2}, which turn up as rates all the time: a swinging weight’s velocity is a sine, a population’s growth rate is an exponential.
Nothing new has to be invented. Every derivative rule already learned, read from right to left, is an antiderivative rule. The derivatives of the six trig functions, of exponentials, of the natural log and of the inverse trig functions together give a table of basic antiderivatives. With the sum and constant multiple rules, that table covers a large share of the integrals met before the special techniques of later lessons.
Definition
Day 1: reading the table backward
As in the power lesson, F is an antiderivative of f on an interval I when F'(x) = f(x) for every x in I, and
names the whole family. A basic antiderivative formula is a derivative rule F' = f written in this form. For example, \dfrac{d}{dx}\big[\tan x\big] = \sec^2 x becomes \int \sec^2 x\,dx = \tan x + C.
The absolute value inside \ln\lvert x \rvert makes the logarithm defined for every x \ne 0: for x > 0 it is \ln x, and for x < 0 it is \ln(-x).
Day 2: inverse trig forms and rewriting
The integrands \dfrac{1}{1 + x^2} and \displaystyle \dfrac{1}{\sqrt{1 - x^2}} are algebraic, built from powers and roots, yet their antiderivatives are the inverse trig functions \arctan x and \arcsin x. An integrand that matches no line of the table can often be rewritten into one that does, by a trig identity, by splitting a quotient into a sum, or by both. The rewritten integrand is the same function, so it has the same antiderivatives.
Intuition
A derivative rule is a pair: a function and its rate. The table of antiderivatives lists the same pairs with the columns swapped. The only care needed is with signs and constants. Since \dfrac{d}{dx}\big[\cos x\big] = -\sin x, the function whose derivative is \sin x is -\cos x, not \cos x. Since \dfrac{d}{dx}\big[e^{3x}\big] = 3e^{3x}, the function whose derivative is e^{3x} is \dfrac{1}{3}e^{3x}. A guess, a derivative, then a correction by a constant factor: that loop finds most of the formulas below.
For \dfrac{1}{x}, think about slopes. On the right, \ln x climbs ever more gently, with slope \dfrac{1}{x}. The graph of \ln\lvert x \rvert on the left is the mirror image of the right branch, so its slopes are the opposites: where the right branch rises with slope \displaystyle \frac{1}{2} at x = 2, the left branch falls with slope \displaystyle -\frac{1}{2} at x = -2. That is exactly \dfrac{1}{x} at x = -2. Figure 1 shows both branches.
Results
Day 1: trig, exponential and logarithm
The trig rules. Each holds on every interval where both sides are defined.
| Derivative rule | Antiderivative |
|---|---|
| \dfrac{d}{dx}\big[\sin x\big] = \cos x | \int \cos x\,dx = \sin x + C |
| \dfrac{d}{dx}\big[\cos x\big] = -\sin x | \int \sin x\,dx = -\cos x + C |
| \dfrac{d}{dx}\big[\tan x\big] = \sec^2 x | \int \sec^2 x\,dx = \tan x + C |
| \dfrac{d}{dx}\big[\sec x\big] = \sec x\tan x | \int \sec x\tan x\,dx = \sec x + C |
| \dfrac{d}{dx}\big[\cot x\big] = -\csc^2 x | \int \csc^2 x\,dx = -\cot x + C |
| \dfrac{d}{dx}\big[\csc x\big] = -\csc x\cot x | \int \csc x\cot x\,dx = -\csc x + C |
Exponentials. For every base a > 0 with a \ne 1,
A constant inside. For a constant k \ne 0,
One over x. On the interval (0, \infty) and on the interval (-\infty, 0),
On the whole domain x \ne 0 the antiderivatives are exactly the functions equal to \ln\lvert x \rvert + C_1 for x > 0 and \ln\lvert x \rvert + C_2 for x < 0, with C_1 and C_2 chosen separately.
Day 2: inverse trig forms and identities
Arctangent and arcsine. For all x, and on (-1, 1) respectively,
Arccosine gives nothing new. On [-1, 1], \arcsin x + \arccos x = \dfrac{\pi}{2}. So -\arccos x + C is the same family as \arcsin x + C, with the constant shifted by \displaystyle \frac{\pi}{2}.
Identities used for rewriting. Wherever both sides are defined,
The first two come from the Pythagorean identity, the last two are the power-reducing formulas.
Why it is true
The trig table. Each line is a derivative rule from Derivatives of the trigonometric functions, and the family result of Antiderivatives and indefinite integrals supplies the +\,C.
Exponentials. \dfrac{d}{dx}\big[a^x\big] = a^x \ln a. Since a \ne 1, \ln a \ne 0, and dividing by it gives \dfrac{d}{dx}\left[\dfrac{a^x}{\ln a}\right] = a^x. The base e is the case \ln e = 1.
A constant inside. By the chain rule, \dfrac{d}{dx}\big[e^{kx}\big] = ke^{kx}, so \dfrac{d}{dx}\left[\dfrac{1}{k}e^{kx}\right] = e^{kx}. The same factor k appears when \sin kx or \cos kx is differentiated, and dividing by k removes it. Undoing the chain rule for an inside function that is not a constant multiple of x needs substitution.
One over x. For x > 0, \ln\lvert x \rvert = \ln x and its derivative is \dfrac{1}{x}. For x < 0, \ln\lvert x \rvert = \ln(-x), and the chain rule gives
So \ln\lvert x \rvert is an antiderivative of \dfrac{1}{x} on each of the two intervals. On each interval separately, two antiderivatives differ by a constant, but nothing ties the constant on one interval to the constant on the other, because no interval joins them.
Arctangent and arcsine. These are the derivatives found in Derivatives of inverse trigonometric functions by the inverse function rule: \dfrac{d}{dx}\big[\arctan x\big] = \dfrac{1}{1 + x^2} for all x, and \displaystyle \dfrac{d}{dx}\big[\arcsin x\big] = \dfrac{1}{\sqrt{1 - x^2}} for -1 < x < 1.
Arccosine. On (-1, 1), \displaystyle \dfrac{d}{dx}\big[\arccos x\big] = -\dfrac{1}{\sqrt{1 - x^2}}, so g(x) = \arcsin x + \arccos x has g'(x) = 0. By the mean value theorem, g is constant on (-1, 1), and \displaystyle g(0) = 0 + \frac{\pi}{2}. Both functions are continuous at \pm 1, so \displaystyle g = \frac{\pi}{2} on all of [-1, 1]. Figure 2 shows the two curves \arcsin x and -\arccos x, one a vertical shift of the other.
Worked examples
Day 1: trig rules term by term
Find (a) \int \left(4\cos x + 3\sin x\right)dx and (b) \int \left(\sec\theta\tan\theta - 2\csc^2\theta\right)d\theta, and check each by differentiating.
Solution(a) The sum and constant multiple rules let each term be handled alone. The antiderivative of \cos x is \sin x, and that of \sin x is -\cos x:
Check: \dfrac{d}{dx}\big[4\sin x - 3\cos x\big] = 4\cos x + 3\sin x. The minus sign on the cosine is what turns -\sin x back into +\sin x.
(b) The variable is \theta. Read the table: \sec\theta\tan\theta comes from \sec\theta, and \csc^2\theta comes from -\cot\theta, so -2\csc^2\theta comes from 2\cot\theta:
Check: the derivative of \sec\theta + 2\cot\theta is \sec\theta\tan\theta - 2\csc^2\theta. The integrand needs \sin\theta \ne 0 and \cos\theta \ne 0, so the answer holds on each interval between consecutive multiples of \displaystyle \frac{\pi}{2}, such as \displaystyle \left(0, \frac{\pi}{2}\right).
Day 1: exponentials, with a constant inside
Find (a) \int \left(2e^x + 3^x\right)dx, (b) \int e^{x/4}\,dx and (c) \int 6e^{-2t}\,dt.
Solution(a) The exponential e^x is its own antiderivative. For base 3 the derivative of 3^x is 3^x \ln 3, one factor of \ln 3 too many, so divide by it:
Check: 2e^x + \dfrac{3^x \ln 3}{\ln 3} = 2e^x + 3^x.
(b) Guess e^{x/4}. Its derivative, by the chain rule, is \displaystyle \frac{1}{4}e^{x/4}: a factor \displaystyle \frac{1}{4} too small. Multiplying the guess by 4 fixes it, which is the rule with \displaystyle k = \frac{1}{4}, since \displaystyle \frac{1}{k} = 4:
Check: \displaystyle 4 \cdot \frac{1}{4}e^{x/4} = e^{x/4}.
(c) Here k = -2, so \displaystyle \int e^{-2t}\,dt = -\frac{1}{2}e^{-2t} + C, and the constant 6 multiplies through:
Check: \dfrac{d}{dt}\big[-3e^{-2t}\big] = -3 \cdot (-2)e^{-2t} = 6e^{-2t}. The antiderivative of a positive function here has a minus sign; differentiating is what confirms it.
Day 1: Arami and Yvoty disagree about one over x
Arami writes \int \dfrac{1}{x}\,dx = \ln x + C, and Yvoty writes \ln\lvert x \rvert + C. (a) Where is each answer correct? (b) Find the function f with f'(x) = \dfrac{1}{x} for every x \ne 0, f(1) = 2 and f(-1) = 5.
Solution(a) Arami’s \ln x is defined only for x > 0, and there it is correct, since its derivative is \dfrac{1}{x}. But \dfrac{1}{x} is also defined for x < 0, where \ln x does not exist. Yvoty’s \ln\lvert x \rvert is defined for every x \ne 0, and the Why it is true section showed that its derivative is \dfrac{1}{x} on both sides. On (0, \infty) the two answers agree; Yvoty’s also covers (-\infty, 0).
(b) The domain is two intervals, so f may carry a different constant on each:
The condition f(1) = 2 is on the right piece: \ln 1 + C_1 = 2, so C_1 = 2. The condition f(-1) = 5 is on the left piece: \ln 1 + C_2 = 5, so C_2 = 5. Thus f(x) = \ln x + 2 for x > 0 and f(x) = \ln(-x) + 5 for x < 0.
Check: on each piece f'(x) = \dfrac{1}{x}, and f(1) = 2, f(-1) = 5. No single formula \ln\lvert x \rvert + C meets both conditions, since it would need C = 2 and C = 5 at once. One condition per piece is needed, and one condition alone fixes only its own piece.
Day 1: splitting a quotient fills the gap
Find \int \dfrac{2x^2 + 5x - 4}{x^2}\,dx.
SolutionThe denominator is one term, so each term of the numerator can be divided by it, for x \ne 0:
The first and last terms are powers, and the power rule handles them: \int 2\,dx = 2x and \int -4x^{-2}\,dx = -4 \cdot \dfrac{x^{-1}}{-1} = \dfrac{4}{x}. The middle term is the case n = -1, which the power rule cannot touch, and the logarithm handles it: \int \dfrac{5}{x}\,dx = 5\ln\lvert x \rvert. Together,
on (0, \infty) and on (-\infty, 0) separately. Check: 2 + \dfrac{5}{x} - \dfrac{4}{x^2}, which is the integrand split up.
Day 2: inverse trig forms, and two answers that agree
Find (a) \int \dfrac{3}{1 + x^2}\,dx, (b) \displaystyle \int \dfrac{2}{\sqrt{1 - x^2}}\,dx, where Jasy gets -2\arccos x + C and Yvoty gets 2\arcsin x + C, and (c) \int \dfrac{x^2}{1 + x^2}\,dx.
Solution(a) The constant comes out and the rest is the arctangent form:
This holds for all x. Check: 3 \cdot \dfrac{1}{1 + x^2}.
(b) Both are right, on (-1, 1). Yvoty read the arcsine line: \displaystyle \dfrac{d}{dx}\big[2\arcsin x\big] = \dfrac{2}{\sqrt{1 - x^2}}. Jasy remembered that arccosine has the opposite derivative: the derivative of -2\arccos x is \displaystyle -2 \cdot \left(-\dfrac{1}{\sqrt{1 - x^2}}\right), which is \displaystyle \dfrac{2}{\sqrt{1 - x^2}}. Since \displaystyle \arcsin x + \arccos x = \frac{\pi}{2},
Yvoty’s function is Jasy’s plus \pi. The two families are the same set of functions, with constants that differ by \pi, as Figure 2 shows for the unscaled curves.
(c) The integrand matches no line of the table, but its numerator is close to its denominator. Adding and subtracting 1 splits the quotient:
Now each term is in the table:
Check: 1 - \dfrac{1}{1 + x^2} = \dfrac{x^2}{1 + x^2}.
Day 2: an identity first
Find (a) \int \tan^2 x\,dx, (b) \int \sin^2 x\,dx and (c) \int \dfrac{\sin x}{\cos^2 x}\,dx.
SolutionNone of the three appears in the table as written. Each is rewritten into lines that do.
(a) By the Pythagorean identity, \tan^2 x = \sec^2 x - 1, and both terms are in the table:
on each interval where \cos x \ne 0. Check: \sec^2 x - 1 = \tan^2 x. Writing \dfrac{\tan^3 x}{3} instead fails the check, because by the chain rule its derivative is \tan^2 x \sec^2 x.
(b) The power-reducing formula turns a square into a first power with a constant inside:
The \displaystyle \frac{1}{4} is \displaystyle \frac{1}{2} from the identity times \displaystyle \frac{1}{2} from the constant k = 2 inside. Check: the derivative is \dfrac{1}{2} - \dfrac{2\cos 2x}{4}, which is \dfrac{1 - \cos 2x}{2}, that is, \sin^2 x.
(c) Split the denominator into two factors:
So \int \dfrac{\sin x}{\cos^2 x}\,dx = \sec x + C on each interval where \cos x \ne 0. Check: \dfrac{d}{dx}\big[\sec x\big] = \sec x\tan x = \dfrac{\sin x}{\cos^2 x}.
Day 2: initial conditions with exponentials and trig
(a) Find f if f'(x) = 3e^x + \dfrac{4}{1 + x^2} and f(0) = 5. (b) Jasy follows a particle moving along a line. Its acceleration at time t seconds is a(t) = -8\sin 2t centimeters per second per second; at t = 0 its velocity is 4 centimeters per second and its position is 0. Find its velocity v(t) and position s(t), and describe the motion.
Solution(a) Term by term, f(x) = 3e^x + 4\arctan x + C. At x = 0, e^0 = 1 and \arctan 0 = 0, so 3 + 0 + C = 5 and C = 2:
Check: f'(x) = 3e^x + \dfrac{4}{1 + x^2} and f(0) = 3 + 0 + 2 = 5.
(b) Velocity is an antiderivative of acceleration. With k = 2, \displaystyle \int \sin 2t\,dt = -\frac{1}{2}\cos 2t + C, so
Then v(0) = 4\cos 0 + C_1 = 4 + C_1 = 4 gives C_1 = 0, and v(t) = 4\cos 2t. Position is an antiderivative of velocity:
and s(0) = 0 + C_2 = 0 gives s(t) = 2\sin 2t centimeters. Check: s'(t) = 4\cos 2t = v(t) and v'(t) = -8\sin 2t = a(t).
The particle swings back and forth between s = -2 and s = 2 centimeters, repeating every \pi seconds, the period of \sin 2t. Had the starting velocity been different, C_1 would not be 0 and the term C_1 t in s(t) would make the particle drift steadily along the line as it swings.
Boundaries
- Each formula holds on an interval. \tan x, \sec x, \cot x and \csc x are undefined at isolated points, and \ln\lvert x \rvert at 0. An answer such as \tan x - x + C is read on each interval of the domain separately, and on a domain made of several intervals each piece may carry its own constant (Example 3).
- The base of a^x must be positive and not 1. For a = 1 the formula would divide by \ln 1 = 0; the integrand is then the constant 1, with antiderivative x + C.
- A constant inside means kx, nothing more. The rule \displaystyle \int e^{kx}\,dx = \frac{1}{k}e^{kx} + C divides by a constant. For e^{x^2}, dividing by 2x fails the check: the quotient rule gives the derivative of \dfrac{e^{x^2}}{2x} as e^{x^2} - \dfrac{e^{x^2}}{2x^2}, not e^{x^2}. In fact e^{x^2} has no antiderivative built from familiar functions. An inside function such as x^2 + 1 with its derivative present as a factor is the job of u-substitution.
- The arcsine form needs -1 < x < 1. Outside that interval \displaystyle \sqrt{1 - x^2} is not a real number. A nearby integrand such as \dfrac{1}{1 - x^2}, with no root, is a different function, and its antiderivative comes from partial fractions. Forms such as \dfrac{1}{4 + x^2} or \dfrac{1}{x^2 + 2x + 2} are reached by substitution and by completing the square.
- Angles in radians. Every trig formula here uses the radian derivative \dfrac{d}{dx}\big[\sin x\big] = \cos x. With x in degrees an extra factor \displaystyle \frac{\pi}{180} appears.
- No quotient or product rule in reverse. \int \dfrac{\sin x}{\cos^2 x}\,dx is not an antiderivative of the top divided by an antiderivative of the bottom: \displaystyle \dfrac{-\cos x}{\frac{x}{2} + \frac{1}{4}\sin 2x} does not differentiate back to it. It was found by rewriting (Example 6).
Where this goes next
u-substitution undoes the chain rule for inside functions beyond kx, so that \int 2x\cos x^2\,dx and \int \dfrac{3x^2}{x^3 + 1}\,dx become one-line problems resting on this table. Trigonometric integrals extends Example 6 to higher powers of sine, cosine, secant and tangent. Dividing before integrating generalizes Example 4 to denominators with several terms, often leaving a \ln or an \arctan behind, and Completing the square inside an integral reshapes quadratic denominators into the arctangent form. With the fundamental theorem, every formula here also finds areas, by The second fundamental theorem of calculus.
Exercises
Day 1: the trig table
- Find \int \left(2\sin x + 7\cos x\right)dx.
- Find \int 5\csc^2 t\,dt.
- Find \int \left(\sec^2 x - 3\sec x\tan x\right)dx.
- Find \int \left(6\csc\theta\cot\theta + \theta^2\right)d\theta.
- Decide whether F is an antiderivative of f: (a) F(x) = \sin^2 x and f(x) = 2\sin x\cos x; (b) F(x) = \cot x and f(x) = \csc^2 x.
Day 1: exponentials and one over x
- Find \int \left(4e^x - x^3\right)dx.
- Find \int 7^x\,dx.
- Find \int e^{5x}\,dx.
- Find \int 8e^{-x/2}\,dx.
- Find \int \cos 3x\,dx.
- Find \int \dfrac{6}{x}\,dx.
- Find \int \dfrac{x^2 + 4}{x}\,dx.
- Find \int \left(\dfrac{3}{x} - 2\sin x + e^x\right)dx.
- Find \int \left(\dfrac{1}{2}\right)^x dx.
Day 2: inverse trig forms and rewriting
- Find \int \dfrac{5}{1 + t^2}\,dt.
- Find \displaystyle \int \dfrac{1}{3\sqrt{1 - x^2}}\,dx.
- Find \int \dfrac{x^2 + 2}{x^2 + 1}\,dx.
- Find \int \cot^2 x\,dx.
- Find \int \cos^2 x\,dx.
- Find \int \dfrac{\cos x}{\sin^2 x}\,dx.
Day 2: initial conditions
- Find f if f'(x) = 2\cos x - 3\sin x and \displaystyle f\left(\frac{\pi}{2}\right) = 6.
- Find f on (-\infty, 0) if f'(x) = \dfrac{4}{x} and f(-e) = 1.
- Find f if f''(x) = e^x - \cos x, f'(0) = 3 and f(0) = 0.
Reasoning
- For x > 0, Arami says \int \dfrac{1}{3x}\,dx = \dfrac{1}{3}\ln 3x + C, and Yvoty says it is \dfrac{1}{3}\ln x + C. Show that both are right, and find the constant by which the two functions differ.
- Show that \dfrac{1}{2}\sin^2 x and -\dfrac{1}{4}\cos 2x are both antiderivatives of \sin x\cos x, and explain why the two answers do not conflict.
Answers
- -2\cos x + 7\sin x + C. The antiderivative of \sin x is -\cos x and that of \cos x is \sin x.
- -5\cot t + C. Since \dfrac{d}{dt}\big[\cot t\big] = -\csc^2 t, the antiderivative of \csc^2 t is -\cot t.
- \tan x - 3\sec x + C, on each interval where \cos x \ne 0. Read the \sec^2 and \sec\tan lines.
- -6\csc\theta + \dfrac{\theta^3}{3} + C, on each interval where \sin\theta \ne 0. The antiderivative of \csc\theta\cot\theta is -\csc\theta.
- (a) Yes. By the chain rule F'(x) = 2\sin x \cdot \cos x. (b) No. F'(x) = -\csc^2 x; an antiderivative of f is -\cot x.
- 4e^x - \dfrac{x^4}{4} + C. Term by term.
- \dfrac{7^x}{\ln 7} + C. The derivative of 7^x is 7^x \ln 7, so divide by \ln 7.
- \dfrac{1}{5}e^{5x} + C. The rule with k = 5.
- -16e^{-x/2} + C. With \displaystyle k = -\frac{1}{2}, \displaystyle \frac{1}{k} = -2, and 8 \cdot (-2) = -16.
- \dfrac{1}{3}\sin 3x + C. The rule with k = 3; check: \displaystyle \frac{1}{3} \cdot 3\cos 3x.
- 6\ln\lvert x \rvert + C, on each side of 0.
- \dfrac{x^2}{2} + 4\ln\lvert x \rvert + C. Split: \dfrac{x^2 + 4}{x} = x + \dfrac{4}{x} for x \ne 0.
- 3\ln\lvert x \rvert + 2\cos x + e^x + C. The antiderivative of -2\sin x is 2\cos x.
- -\dfrac{1}{\ln 2}\left(\dfrac{1}{2}\right)^x + C. The base is \displaystyle a = \frac{1}{2} and \displaystyle \ln\frac{1}{2} = -\ln 2.
- 5\arctan t + C. The arctangent form times 5.
- \dfrac{1}{3}\arcsin x + C, on (-1, 1). The arcsine form times \displaystyle \frac{1}{3}.
- x + \arctan x + C. Split: \dfrac{x^2 + 2}{x^2 + 1} is \dfrac{(x^2 + 1) + 1}{x^2 + 1}, which is 1 + \dfrac{1}{1 + x^2}.
- -\cot x - x + C, on each interval where \sin x \ne 0. Rewrite \cot^2 x = \csc^2 x - 1.
- \dfrac{x}{2} + \dfrac{1}{4}\sin 2x + C. Rewrite \displaystyle \cos^2 x = \frac{1}{2} + \frac{1}{2}\cos 2x, then use k = 2.
- -\csc x + C, on each interval where \sin x \ne 0. Rewrite \dfrac{\cos x}{\sin^2 x} as \dfrac{1}{\sin x} \cdot \dfrac{\cos x}{\sin x}, which is \csc x\cot x.
- f(x) = 2\sin x + 3\cos x + 4. The family is 2\sin x + 3\cos x + C, and at \displaystyle \frac{\pi}{2} it is 2 + 0 + C = 6.
- f(x) = 4\ln(-x) - 3 for x < 0. The family on (-\infty, 0) is 4\ln(-x) + C, and 4\ln e + C = 4 + C = 1.
- f(x) = e^x + \cos x + 2x - 2. First f'(x) = e^x - \sin x + C_1 with 1 - 0 + C_1 = 3, so C_1 = 2. Then f(x) = e^x + \cos x + 2x + C_2 with 1 + 1 + C_2 = 0, so C_2 = -2.
- Both differentiate to \dfrac{1}{3x}: by the chain rule the derivative of \dfrac{1}{3}\ln 3x is \dfrac{1}{3} \cdot \dfrac{3}{3x}, which is \dfrac{1}{3x}, and \dfrac{d}{dx}\left[\dfrac{1}{3}\ln x\right] = \dfrac{1}{3x}. By the product rule for logarithms, \ln 3x = \ln 3 + \ln x, so Arami’s function is Yvoty’s plus the constant \dfrac{\ln 3}{3}.
- The derivative of \dfrac{1}{2}\sin^2 x is \sin x\cos x by the chain rule, and the derivative of -\dfrac{1}{4}\cos 2x is \dfrac{1}{2}\sin 2x, which is \sin x\cos x by the double-angle formula. They do not conflict because they differ by a constant: \cos 2x = 1 - 2\sin^2 x gives -\dfrac{1}{4}\cos 2x = \dfrac{1}{2}\sin^2 x - \dfrac{1}{4}.