1The idea in one paragraph

Every new derivative on this page comes from a rule you already know. tan, sec, cosec and cot are quotients or reciprocals of sin and cos, so the quotient rule or the chain rule differentiates them. aˣ is really e^{x ln a}, and logₐ x is really ln x ÷ ln a. arcsin, arccos and arctan are inverse functions, so writing y = arcsin x as sin y = x and differentiating implicitly gives their derivatives. Then read every result backwards: each derivative is an integral you can now do. The two that matter most are 1/(a² + x²), whose integral is an arctan, and 1/√(a² − x²), whose integral is an arcsin. Completing the square turns many quadratic denominators into those shapes, and partial fractions turn the rest into logarithms.

2Four more trigonometric derivatives

tan x. Write it as sin x ÷ cos x and use the quotient rule.

sec x. Write it as (cos x)⁻¹ and use the chain rule.

cosec x and cot x come out the same way, from (sin x)⁻¹ and cos x ÷ sin x. All four are in the formula booklet.

A pattern makes them easy to hold. Swap each function for its "co-" partner (tan ↔ cot, sec ↔ cosec) and put a minus sign in front: tan → sec² becomes cot → −cosec², and sec → sec tan becomes cosec → −cosec cot. The minus sign belongs to the co-functions, just as d/dx(cos x) = −sin x.

Figure 1 shows what the first result means. sec²x is never less than 1, so tan x is increasing on every branch, and its gentlest slope is 1, at x = 0. Near its asymptotes the gradient grows without bound.

With the chain rule, the inner derivative multiplies the front, exactly as in SL.

3aˣ and logₐ x

aˣ. Any positive base can be rewritten with base e, because a = e^{ln a}. So aˣ = e^{x ln a}, and the chain rule does the rest.

d/dx(aˣ) = aˣ ln a and d/dx(logₐ x) = 1/(x ln a), both in the formula booklet.

Figure 2 draws 2ˣ, eˣ and 3ˣ with their tangents at x = 0. Each passes through (0, 1), and the gradient there is ln a: ln 2 ≈ 0.693 for 2ˣ, ln e = 1 for eˣ, ln 3 ≈ 1.10 for 3ˣ. That is why e is the natural base: it is the one base whose exponential has gradient exactly 1 at x = 0, so the ln a factor disappears.

logₐ x. Use the change of base law from SL 1.7: logₐ x = ln x ÷ ln a. Since ln a is only a constant, d/dx(logₐ x) = (1/ln a) × (1/x) = 1/(x ln a).

The third line surprises people. It makes sense once you split the log: log₁₀(5x) = log₁₀ 5 + log₁₀ x, and log₁₀ 5 is a constant, so it differentiates to zero.

4arcsin x, arccos x and arctan x

These come from implicit differentiation (AHL 5.14). The guide expects you to be able to derive them, and a "show that" on one of them is a standard 3 to 4 marks.

arcsin x. Let y = arcsin x, so sin y = x, with −π/2 ≤ y ≤ π/2.

The reason for the positive root is worth a line in your answer: arcsin's range is [−π/2, π/2] (AHL 3.9), where cos is never negative.

arccos x. Let y = arccos x, so cos y = x with 0 ≤ y ≤ π. Then −sin y · dy/dx = 1, and sin y = +√(1 − x²) on [0, π], so d/dx(arccos x) = −1/√(1 − x²). It is the negative of the arcsin derivative, which makes sense: arcsin x + arccos x = π/2, a constant, so their derivatives must add to zero.

arctan x. Let y = arctan x, so tan y = x.

Figure 3 puts arcsin and arctan beside their derivatives. arcsin climbs steeply at x = ±1, where its derivative has vertical asymptotes; arctan is steepest at 0, where its derivative reaches its maximum of 1, and flattens towards its asymptotes y = ±π/2, where its derivative tends to 0.

With the chain rule:

5Reading the derivatives backwards

Every derivative gives an integral for free. The guide asks for the indefinite integrals of all of them.

The formula booklet gives the last two in a more useful form, with a constant a:

∫ 1/(a² + x²) dx = (1/a) arctan(x/a) + C and ∫ 1/√(a² − x²) dx = arcsin(x/a) + C, for |x| < a.

You can check either by differentiating. For the first: d/dx[(1/a) arctan(x/a)] = (1/a) × (1/a) × 1/(1 + x²/a²) = 1/(a² + x²). Notice the arctan result has a factor 1/a in front and the arcsin result does not. That asymmetry is the most common slip in this subtopic.

The integral of −1/√(1 − x²) could be written −arcsin x + C or arccos x + C. Both are right: they differ by the constant π/2, which the C absorbs.

Two exact definite integrals worth knowing:

Using an identity first. tan²x is not on the list, but 1 + tan²x = sec²x turns it into something that is: ∫ tan²x dx = ∫ (sec²x − 1) dx = tan x − x + C. The same trick handles cot²x with cosec²x.

6A linear function inside

The guide includes the composite of any of these with a linear function, ax + b. The rule is the reverse of the chain rule:

If ∫ f(x) dx = F(x) + C, then ∫ f(ax + b) dx = (1/a) F(ax + b) + C.

Differentiate the answer and the chain rule produces a factor a, which the 1/a cancels. So: integrate as if the bracket were x, then divide by the coefficient of x.

When the coefficient sits on x² rather than x, take it out first so the booklet form appears.

Worked example 1. Paper 1. Find ∫ 1/(16 + 9x²) dx.

Check by differentiating: (1/12) × (3/4) × 1/(1 + 9x²/16) = (1/16) × 16/(16 + 9x²) = 1/(16 + 9x²).

7Completing the square

A quadratic denominator that does not factorise can be written as (x + p)² + q² by completing the square, which is the arctan shape with x + p in place of x. That is a linear composite, so section 6 applies with a = 1.

Worked example 2. Paper 1. Find ∫ 1/(x² − 6x + 13) dx.

The same move works inside a square root, for the arcsin shape. Take out the minus sign carefully.

Worked example 3. Find ∫ 1/√(8x − x²) dx.

Splitting the numerator. When the top is linear, split it into a part that is a multiple of the derivative of the bottom, which integrates to a logarithm (the reverse chain rule of SL 5.10), and a constant part, which gives an arctan.

Worked example 4. Find ∫ (2x + 5)/(x² + 9) dx.

8Partial fractions: when the denominator does factorise

If the quadratic in the denominator factorises, there is no arctan. Instead, split the fraction into partial fractions (AHL 1.11), and each piece integrates to a natural logarithm, because ∫ 1/(x + k) dx = ln|x + k| + C.

Worked example 5. Paper 1. Find ∫ (5x + 1)/((x − 1)(x + 2)) dx.

Worked example 6. Compare two integrals that differ by one sign. With a plus sign the denominator does not factorise and the answer is an arctan: ∫ 1/(x² + 4) dx = (1/2) arctan(x/2) + C. With a minus sign it does, so the answer is logarithms.

The discriminant decides which road you take, and Figure 4 lays out the choice.

9An indefinite integral is a family of curves

∫ f(x) dx = F(x) + C does not describe one function. It describes infinitely many, one for each value of C, and their graphs are vertical translations of each other. Every member has the same gradient at the same x, which is exactly what "the derivative is f(x)" says. Figure 5 draws some members of the family y = arctan 2x + C.

One extra fact, such as a point the curve passes through, picks out a single member.

Worked example 7. Paper 1. A curve has gradient function dy/dx = 2/(1 + 4x²) and passes through (1/2, π). Find its equation.