Difficulty: Intermediate-Advanced | Prerequisites: Part 1 of these notes (inverse derivative formula), chain rule, u-substitution, completing the square.
This is the second half of Unit 7. Now that you understand how to differentiate inverse functions in general, this section applies that idea to the six inverse trigonometric functions and then reverses direction to give you integration formulas. These integration rules are tested heavily on the AP exam, particularly the arcsin and arctan forms. Completing the square is an algebraic prerequisite that reappears here because it lets you reshape ugly integrands into forms that match the standard inverse trig patterns. If you are comfortable with u-substitution and the chain rule, you are ready for this material.
There are six inverse trig derivative formulas, but arcsin, arctan, and arcsec are the ones you need to know cold. Flipping those derivatives around gives three integration rules. Most exam problems test whether you can recognise which rule applies, set up the correct u-substitution, and (sometimes) complete the square first.
Inverse trigonometric function (arc-function)
A function that returns the angle whose trig ratio equals a given value. For example, arcsin(x) returns the angle whose sine is x.
In simple terms, if sin(θ) = x, then arcsin(x) = θ.
u-substitution (in this context)
Rewriting part of the integrand as a new variable u so the integral matches one of the standard inverse trig forms.
In simple terms, you relabel a messy piece of the integral to make it look like a formula you already know.
Completing the square
An algebraic technique that rewrites ax² + bx + c as a(x – h)² + k, making the expression match the a² + u² or a² – u² patterns needed for inverse trig integration.
In simple terms, you rearrange a quadratic so it looks like "something squared plus a constant," which is exactly what the arctan formula wants.
All six formulas below use u as a differentiable function of x. u' denotes du/dx (chain rule).
d/dx [arcsin u] = u' / √(1 – u²)
d/dx [arccos u] = –u' / √(1 – u²)
d/dx [arctan u] = u' / (1 + u²)
d/dx [arccot u] = –u' / (1 + u²)
d/dx [arcsec u] = u' / (|u| √(u² – 1))
d/dx [arccsc u] = –u' / (|u| √(u² – 1))
Notice the pattern: cosine-family derivatives (arccos, arccot, arccsc) are the negatives of their sine-family counterparts (arcsin, arctan, arcsec).
d/dx [arcsin(2x)] = 2 / √(1 – 4x²). Here u = 2x, so u' = 2, and u² = 4x².
d/dx [arcsin(√x)] = (1/2)x⁻¹ᐟ² / √(1 – x) = 1 / (2√x · √(1 – x)) = 1 / (2√(x(1 – x))). Here u = √x, so u' = 1/(2√x).
These are the derivative formulas read in reverse. Three main forms appear on the AP exam:
∫ du / √(a² – u²) = arcsin(u/a) + C
∫ du / (a² + u²) = (1/a) arctan(u/a) + C
∫ du / (u√(u² – a²)) = (1/a) arcsec(|u|/a) + C (rarely tested on the AP exam)
The first two are by far the most important.
∫ dx / √(4 – x²)
Identify a² = 4, so a = 2, and u = x.
Apply the arcsin rule: arcsin(x/2) + C.
∫ (1/3) · 3 dx / (2 + 9x²)
Rewrite 2 + 9x² as (√2)² + (3x)². So a = √2, u = 3x, du = 3 dx.
The integral becomes (1/3) ∫ du / (a² + u²).
Apply the arctan rule: (1/3) · (1/√2) arctan(3x/√2) + C.
∫ (x + 2) / √(4 – x²) dx
Split into two integrals: ∫ x / √(4 – x²) dx + ∫ 2 / √(4 – x²) dx.
The first piece is a standard u-substitution (let u = 4 – x², du = –2x dx), giving –√(4 – x²).
The second piece is the arcsin rule, giving 2 arcsin(x/2).
Final answer: –√(4 – x²) + 2 arcsin(x/2) + C.
∫ dx / (x² – 4x + 7)
Complete the square in the denominator: x² – 4x + 7 = x² – 4x + 4 + 3 = (x – 2)² + 3.
Now identify a² = 3 (so a = √3) and u = x – 2 (so du = dx).
Apply the arctan rule: (1/√3) arctan((x – 2)/√3) + C.
The three AP-critical integration formulas:
∫ du / √(a² – u²) = arcsin(u/a) + C
∫ du / (a² + u²) = (1/a) arctan(u/a) + C
∫ du / (u√(u² – a²)) = (1/a) arcsec(|u|/a) + C
Recognition guide:
Square root of (a² minus u²) in the denominator → arcsin
Sum of squares (a² plus u²) in the denominator, no square root → arctan
u times square root of (u² minus a²) in the denominator → arcsec
Inverse trig integrals appear in physics when computing the time for an object under certain force laws, in electrical engineering when analysing RC and RLC circuit responses, and in probability when working with distributions whose densities involve 1/(1 + x²) (the Cauchy distribution). The arctan integral in particular shows up any time a quantity approaches a limit asymptotically.
Students often forget to factor out 1/a in the arctan result. The integral of du/(a² + u²) is (1/a) arctan(u/a), not just arctan(u/a). Missing that leading 1/a is one of the most common point losses.
Confusing the arcsin and arctan forms. If the denominator has a square root, you are probably looking at arcsin. If there is no square root, it is probably arctan. Train yourself to check for the radical first.
Trying to apply the arcsin formula when the form is √(u² – a²) rather than √(a² – u²). The sign arrangement matters: a² – u² goes with arcsin, u² – a² goes with arcsec.
Forgetting to complete the square. If the denominator is a quadratic that does not already look like a² + u², you almost certainly need to complete the square first. Students who skip this step get stuck.
⚠️ The arcsin and arctan integration formulas appear on nearly every AP Calc AB/BC exam, either in multiple choice or as part of a free-response integration problem.
⚠️ Completing the square combined with arctan integration is a classic "hard" multiple-choice question. Practise it until the algebra is automatic.
⚠️ Splitting an integrand into a u-sub piece and an inverse trig piece (as in Example 3) is a frequent exam technique. Look for a sum in the numerator as the clue.
⚠️ The arcsec integration rule is noted in these lecture notes as "rarely seen on AP." Prioritise arcsin and arctan.
Fill in the blank: d/dx [arctan(x)] = ______.
True or false: ∫ du / (a² + u²) = arctan(u/a) + C.
Fill in the blank: When the denominator contains √(a² – u²), the corresponding inverse trig integral produces ______.
True or false: To integrate dx / (x² – 6x + 13), you should complete the square first.
What is d/dx [arcsin(3x)]?
Q: Evaluate ∫ dx / √(9 – x²).
A: Here a = 3, u = x. Apply the arcsin formula: arcsin(x/3) + C.
Q: Evaluate ∫ dx / (1 + 4x²).
A: Rewrite as ∫ dx / (1 + (2x)²). Let u = 2x, du = 2 dx, so dx = du/2. The integral becomes (1/2) ∫ du / (1 + u²) = (1/2) arctan(u) + C = (1/2) arctan(2x) + C.
Q: Evaluate ∫ dx / (x² + 6x + 13).
A: Complete the square: x² + 6x + 13 = (x + 3)² + 4. Now a = 2, u = x + 3. Apply arctan: (1/2) arctan((x + 3)/2) + C.
Q: What is d/dx [arccos(x)] and how does it relate to d/dx [arcsin(x)]?
A: d/dx [arccos(x)] = –1/√(1 – x²). It is the negative of d/dx [arcsin(x)] = 1/√(1 – x²). This is because arcsin(x) + arccos(x) = π/2 for all x in [–1, 1], so their derivatives must sum to zero.
Q: Evaluate ∫ (x + 3) / √(1 – x²) dx.
A: Split: ∫ x/√(1 – x²) dx + 3 ∫ 1/√(1 – x²) dx. The first integral uses u-sub (u = 1 – x²) to give –√(1 – x²). The second is 3 arcsin(x). Final answer: –√(1 – x²) + 3 arcsin(x) + C.
The inverse trig integration rules connect back to the general inverse derivative formula from Part 1, since each rule is derived by differentiating the corresponding arc-function. These formulas also set the stage for integration by parts and partial fractions in later units, where you may need to combine inverse trig rules with other techniques. In BC Calculus, these same patterns reappear in improper integrals and series involving arctangent.
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