Integrating eˣ, 1/x, sin x, cos x and sec² x, and sums and multiples of them

Why the integral of 1/x is ln|x|, with a modulus sign

Integrating functions of ax+b: (ax+b)ⁿ, 1/(ax+b), eᵃˣ⁺ᵇ, sin(ax+b) and cos(ax+b)

Integrating by recognition: f′(x)/f(x) gives ln|f(x)|, including ∫tan x dx

Integrating by recognition: f′(x)[f(x)]ⁿ gives a power of f(x)

Using identities to integrate sin² x, cos² x and tan² x

Exact definite integrals and areas, including regions below the x-axis and between two curves

1. Every standard integral in P3 is a P3 derivative read backwards, so ∫eˣ dx = eˣ + c, ∫1/x dx = ln|x| + c, ∫sin x dx = -cos x + c and ∫cos x dx = sin x + c, and you can check any answer by differentiating it.

2. Linear inside? Divide by a: ∫f(ax+b) dx = 1/aF(ax+b) + c. This only works when the inside is linear.

3. Top is (a multiple of) the derivative of the bottom? ∫f′(x)/f(x) dx = ln|f(x)| + c; adjust any constant factor.

4. A power of a function times the derivative of its inside? ∫f′(x)[f(x)]ⁿ dx = [f(x)]ⁿ⁺¹/(n+1) + c.

5. Squares of trigonometric functions must be rewritten first: sin² x = ½(1-cos 2x), cos² x = ½(1+cos 2x), tan² x = sec² x - 1.

6. A region below the x-axis gives a negative integral, so for an area you split at the roots and add the sizes.

Integration is the reverse of differentiation. In P3 you met the derivatives of eˣ, ln x, sin x, cos x and tan x. Read each of those results from right to left and you have a standard integral. That is the whole idea of this page: before you integrate anything, ask which derivative it came from.

The result for 1/x fills the one gap in the P1 power rule. The rule ∫xⁿ dx = xⁿ⁺¹/(n+1) + c breaks down when n = -1, because you would divide by zero.

The function x⁻¹ is the only power that does not integrate to a power; it integrates to a logarithm instead.

Why is there a modulus in ln|x|?

The function ln x only exists for x > 0, but y = 1/x has a second branch for x < 0. On that branch, differentiate ln(-x) with the chain rule: d/dxln(-x) = (-1)/(-x) = 1/x. So ln(-x) is also an antiderivative of 1/x, valid for negative x. Writing ln|x| covers both branches in one expression.

Each trigonometric result needs radians.

If x were measured in degrees, the derivative of sin x would carry a factor π/180, and so would every integral. Keep your calculator in radian mode for the whole topic.

WORKED EXAMPLE · QUESTION

(a) Find ∫(4eˣ - 3/x + 5sin x - 2cos x)dx, x > 0. (b) Evaluate ∫₀^(π/2) (5sin x - 2cos x) dx.

SOLUTION

(a) Integrate term by term, watching the signs on the trigonometric terms: 4eˣ stays 4eˣ; -3/x gives -3ln x; 5sin x gives -5cos x; -2cos x gives -2sin x. Because x > 0, ln|x| can be written ln x. 4eˣ - 3ln x - 5cos x - 2sin x + c

(b) Use the antiderivative -5cos x - 2sin x and substitute the limits: (-5×0 - 2×1) - (-5×1 - 2×0) = -2 + 5. 3

By the chain rule, d/dx[F(ax+b)] = a F′(ax+b). The derivative picks up an extra factor a, the coefficient of x. Reversing this, you integrate as if the bracket were a single x, then divide by a to cancel the factor the chain rule would put back.

There is a neat picture behind the factor 1/a. The graph of y = cos 2x is the graph of y = cos x squashed horizontally by a factor of 2.

Every vertical strip becomes half as wide but keeps its height, so every area is halved. That is exactly the ½ in ∫cos 2x dx = ½sin 2x + c.

The rule only works when the inside is linear. For ∫(x²+1)³ dx the derivative of the inside is 2x, which is not a constant, so you cannot simply divide by it. Either expand the bracket or, if a factor of x is present outside, use the pattern in the section on f′(x)[f(x)]ⁿ.

WORKED EXAMPLE · QUESTION

Find (a) ∫(3x-2)⁴ dx (b) ∫8/(2x+5) dx (c) ∫e¹⁻⁴ˣ dx (d) ∫sin(2x + π/6)dx.

SOLUTION

(a) Raise the power to 5, divide by the new power and by a = 3: ((3x-2)⁵)/(5×3). (1/15)(3x-2)⁵ + c

(b) 8 ×½ln|2x+5|. 4ln|2x+5| + c

(c) Here a = -4, so divide by -4. -¼ e¹⁻⁴ˣ + c

(d) ∫sin u du = -cos u, then divide by 2. -½cos(2x + π/6) + c

WORKED EXAMPLE · QUESTION

Find the exact value of (a) ∫₁⁴ 6/(2x+1) dx (b) ∫₀^(ln 2) e³ˣ dx.

SOLUTION

(a) Integrate, dividing by 2: [3ln|2x+1|]₁⁴ = 3ln 9 - 3ln 3. Combine with the laws of logarithms: 3ln(9/3) = 3ln 3 (≈ 3.30). 3ln 3

(b) [⅓ e³ˣ]₀^(ln 2). Use e^(3ln 2) = e^(ln 8) = 8: ⅓(8 - 1). (7/3)

By the chain rule, d/dxln|f(x)| = f′(x)/f(x). So whenever the numerator of a fraction is the derivative of its denominator, or a constant multiple of it, the integral is a logarithm. This is the most useful recognition pattern on the paper, and the test is quick: differentiate the bottom and compare it with the top.

The most famous case is tan x. Write it as sin x/cos x. The derivative of cos x is -sin x, so the numerator is -1 times the derivative of the denominator:

The last step uses -ln|cos x| = ln|cos x|⁻¹ = ln|sec x|. Similarly ∫cot x dx = ∫cos x/sin x dx = ln|sin x| + c.

The numerator must be a constant multiple of the derivative of the denominator. ∫x/(x²+1) dx = ½ln(x²+1) + c works, because x is half of 2x. But ∫1/(x²+1) dx does not fit the pattern at all: the top would need to contain x.

WORKED EXAMPLE · QUESTION

(a) Find ∫(4x-2)/(x²-x+3) dx. (b) Show that ∫₀^(π/2) cos x/(3+sin x) dx = ln(4/3).

SOLUTION

(a) Differentiate the denominator: d/dx(x²-x+3) = 2x-1, and 4x-2 = 2(2x-1). The integrand is twice f′/f, so it integrates to twice a logarithm. No modulus is needed: x²-x+3 has discriminant 1 - 12 < 0, so it is always positive. 2ln(x²-x+3) + c

(b) The numerator cos x is the derivative of 3 + sin x, which is always positive. So [ln(3+sin x)]₀^(π/2) = ln 4 - ln 3. ln(4/3) as required.

The chain rule gives d/dx[f(x)]ⁿ⁺¹/(n+1) = f′(x)[f(x)]ⁿ. So a power of a function, multiplied by the derivative of that function, integrates by raising the power, exactly as if f(x) were a single variable.

The method "guess, differentiate, adjust" is more reliable than trying to remember a formula. You only ever have to fix a constant factor.

If the factor you need is a function of x, the pattern does not apply and you need a different method (substitution, in P4).

WORKED EXAMPLE · QUESTION

(a) Find ∫x²(x³+2)⁴ dx. (b) Evaluate ∫₀^(π/2) sin xcos³ x dx.

SOLUTION

(a) Guess (x³+2)⁵ and differentiate: d/dx(x³+2)⁵ = 5(x³+2)⁴ ×3x² = 15x²(x³+2)⁴. That is 15 times too big, so divide by 15. (1/15)(x³+2)⁵ + c

(b) cos³ x = (cos x)³, and its inside, cos x, has derivative -sin x. Guess cos⁴ x: d/dxcos⁴ x = -4sin xcos³ x, so ∫sin xcos³ x dx = -¼cos⁴ x + c. With limits: [-¼cos⁴ x]₀^(π/2) = 0 - (-¼). ¼

There is no standard integral for sin² x, cos² x or tan² x. The trick is to rearrange an identity so that the square disappears, and then use the earlier sections. The double-angle formula cos 2x = 1 - 2sin² x = 2cos² x - 1 gives the first two, and 1 + tan² x = sec² x gives the third.

The graphs show why these identities are natural. The curve y = cos² x never goes below zero; it oscillates between 0 and 1 about the mean level ½, with period π. That is exactly a cosine wave ½cos 2x lifted up by ½.

WORKED EXAMPLE · QUESTION

(a) Show that ∫₀^(π/4) cos² x dx = (π+2)/8. (b) Find ∫tan² 3x dx.

SOLUTION

(a) Use the identity: ∫₀^(π/4) ½(1+cos 2x) dx. Integrate, dividing the cos 2x term by 2: [x/2 + sin 2x/4]₀^(π/4). With sin π/2 = 1: (π/8 + ¼) - 0. (π+2)/8 as required.

(b) tan² 3x = sec² 3x - 1, and ∫sec² kx dx = 1/ktan kx. ⅓tan 3x - x + c

A useful sanity check: over a full period the average of sin² x is ½, so ∫₀^π sin² x dx = π/2. The formula agrees: [x/2 - sin 2x/4]₀^π = π/2.

The rules for definite integrals are the same as in P2; only the functions are new.

∫ₐᵇ f(x) dx = F(b) - F(a). The constant cancels, so leave it out.

A region below the x-axis gives a negative integral. Find where the curve crosses the axis, integrate each part separately and add the sizes.

The area between two curves is ∫(upper - lower) dx between their intersections; it does not matter whether the region is above or below the axis.

For trigonometric functions use radians and exact values such as cos π/3 = ½.

WORKED EXAMPLE · QUESTION

The curves y = sin 2x and y = sin x meet at the origin. (a) Show that, for 0 < x < π, the curves meet only where x = π/3. (b) Find the exact area of the finite region enclosed by the two curves for 0 ≤x ≤π/3.

SOLUTION

(a) Use sin 2x = 2sin xcos x: 2sin xcos x = sin x ⇒sin x(2cos x - 1) = 0. Since sin x ≠0 for 0 < x < π, cos x = ½, and x = π/3 is the only solution in (0, π). x = π/3 only

(b) sin 2x is the upper curve on (0, π/3) (at x = π/6: √3/2 > ½).

Area = ∫₀^(π/3)(sin 2x - sin x) dx = [-½cos 2x + cos x]₀^(π/3). With cos 2π/3 = -½ and cos π/3 = ½: (¼ + ½) - (-½ + 1) = ¾ - ½. ¼

WORKED EXAMPLE · QUESTION

The curve C has equation y = x - 5 + 4/x, x > 0. (a) Find the x-coordinates of the points where C crosses the x-axis. (b) Show that the area of the finite region bounded by C and the x-axis is (15/2) - 8ln 2. (c) Find, in the form p + qln 2 + rln 3, the total area of the regions between C, the x-axis and the lines x = 1 and x = 6.

SOLUTION

(a) Multiply by x (allowed, since x > 0): x² - 5x + 4 = 0 ⇒(x-1)(x-4) = 0. x = 1 and x = 4

(b) ∫₁⁴(x - 5 + 4/x)dx = [x²/2 - 5x + 4ln x]₁⁴.

With ln 4 = 2ln 2: (8 - 20 + 8ln 2) - (½ - 5) = -(15/2) + 8ln 2, which is negative because the region is below the axis. The area is its size. (15/2) - 8ln 2 (≈ 1.955)

(c) Between 4 and 6 the curve is above the axis: [x²/2 - 5x + 4ln x]₄⁶ = (18 - 30 + 4ln 6) - (8 - 20 + 4ln 4) = 4ln(6/4) = 4ln 3 - 4ln 2 (≈ 1.622). Add the two sizes: (15/2) - 8ln 2 + 4ln 3 - 4ln 2. (15/2) - 12ln 2 + 4ln 3 (≈ 3.58)

Integrating straight from 1 to 6 gives 4ln 6 - (15/2) ≈-0.333: the parts below and above the axis have partly cancelled. That is not an area.

Using your calculator

Many scientific calculators can evaluate a definite integral numerically. Use it to check an exact answer: 3ln 3 ≈3.30 should match ∫₁⁴ 6/(2x+1) dx.

Set radian mode before any trigonometric integral; in degree mode the check will disagree.

A numerical value never replaces the working when a question asks for an exact answer or says "show that".

P3 Differentiation

Every result here is a P3 derivative reversed: the derivatives of eˣ, ln x, sin x and cos x, the chain rule (for f(ax+b), f′/f and f′fⁿ), and the double-angle and Pythagorean identities from P3 trigonometry.

P1 and P2 Integration

The power rule and the constant of integration come from P1; definite integrals, areas and the trapezium rule from P2. This page widens the set of functions you can apply them to.

P4.6 Integration and beyond

Integration by substitution generalises the f′/f and f′fⁿ patterns; integration by parts, partial fractions (P4.1, which often end in ln terms), volumes of revolution and separable differential equations all rely on these standard results. In mechanics, a velocity such as v = 4e^(-0.5t) or 3sin 2t is integrated to give displacement, and in IAL Physics exponential decay and simple harmonic motion lead to the same integrals.

1. Find ∫e^(-x/2) dx.

2. Find the exact value of ∫₂⁶ 1/x dx.

3. Find ∫3x²/(x³+1) dx.

4. Explain why ∫sin² x dx is not (sin³ x)/3 + c, and find it.

5. Find ∫cos xsin⁴ x dx.

6. Find the exact area under y = tan x between x = 0 and x = π/3.

1. Here a = -½, so divide by -½, i.e. multiply by -2: -2e^(-x/2) + c.

2. ln 6 - ln 2 = ln(6/2) = ln 3.

3. The numerator is the derivative of the denominator, so the integral is ln|x³+1| + c.

4. Differentiating (sin³ x)/3 gives sin² xcos x, not sin² x. Use sin² x = ½(1-cos 2x): ∫sin² x dx = x/2 - sin 2x/4 + c.

5. cos x is the derivative of sin x, so this is f′f⁴: (sin⁵ x)/5 + c.

6. tan x > 0 on this interval, so the area is [ln|sec x|]₀^(π/3) = ln 2 - ln 1 = ln 2 (≈ 0.693), because sec π/3 = 1/(cos(π/3)) = 2.