Maths › Further calculus › Calculus with inverse trigonometric functions
Calculus with inverse trigonometric functions
Differentiating arcsin and arctan produces algebraic fractions. Read backwards, that is a gift, because whole families of integrals suddenly have names.
Builds on Implicit differentiation and Reciprocal and inverse trigonometric functions.
IN THIS TOPIC
- Differentiate arcsin x, arccos x and arctan x from scratch via implicit differentiation.
- Integrate 1/√(a² − x²) and 1/(a² + x²) by recognising the patterns.
- Complete the square and use partial fractions to reach the same two patterns.
- Choose and carry out the substitutions x = a sin θ and x = a tan θ for the associated forms.
COMMON MISCONCEPTION
Since arcsin is a trig-flavoured function, its derivative must involve sines and cosines.
Differentiating the inverses
Let y = arcsin x, so sin y = x. Differentiate implicitly to get cos y (dy/dx) = 1. Since cos y = √(1 − sin²y) = √(1 − x²), the trig evaporates entirely:
Both sit in the booklet's Further Maths differentiation table, alongside arccos, so you are never stuck for the answer. Learn the derivation anyway, because a 'show that' wants the working and not the printed line. arccos differs from arcsin only by a sign, since the two angles sum to a right angle. The right triangle below is the whole derivation in one picture.
WORKED EXAMPLE
A gradient with no trig in it
Find the gradient of y = arcsin x at x = 1/2.
dy/dx = 1/√(1 − x²) = 1/√(1 − 1/4) = 1/√(3/4).
= 2/√3 ≈ 1.155.
As x → 1 the denominator tends to 0, so the graph steepens to vertical at the edges of its domain, exactly as the flipped sine curve suggests.
Reading the table backwards
Every derivative is an integral in reverse. Scaling x by a stretches the results into the exam's two standard integrals, ∫1/√(a² − x²) dx = arcsin(x/a) + c and ∫1/(a² + x²) dx = (1/a) arctan(x/a) + c. Spot the form, name a, write the answer.
WORKED EXAMPLE
π from an algebraic fraction
Evaluate ∫ 1/(9 + x²) dx from 0 to 3.
Here a = 3, so the integral is (1/3) arctan(x/3).
At the limits: (1/3)(arctan 1 − arctan 0) = (1/3)(π/4).
= π/12. No circle in sight, yet π appears, because arctan carries it in.
GUIDED PRACTICE
An arcsin integral
Evaluate ∫ 1/√(4 − x²) dx from 0 to 1.
Show the working
a = 2, so the integral is arcsin(x/2).
arcsin(1/2) − arcsin 0 = π/6 − 0.
The answer is π/6, exact. A decimal would throw the mark away.
Getting an integral into that shape
Most exam integrands need dressing first. Two moves do nearly all of it. Complete the square when the denominator is a general quadratic, so ∫1/(x² + 2x + 5) dx becomes ∫1/((x + 1)² + 4) dx = ½arctan((x + 1)/2) + c. Split the numerator when it contains an x, since ∫x/(x² + 4) dx is a logarithm by recognition while ∫1/(x² + 4) dx is an arctan; keeping them together produces neither.
WORKED EXAMPLE
Partial fractions with a quadratic factor
Find ∫ (x + 4)/((x + 1)(x² + 4)) dx.
The quadratic factor needs a linear numerator, so write (x + 4)/((x + 1)(x² + 4)) = A/(x + 1) + (Bx + C)/(x² + 4).
Multiplying up, x + 4 = A(x² + 4) + (Bx + C)(x + 1). Putting x = −1 gives 5A = 3, and comparing x² and constant terms gives B = −3/5, C = 8/5.
The second fraction splits again, into −(3/5)x/(x² + 4) and (8/5)/(x² + 4).
Integrating: (3/5)ln|x + 1| − (3/10)ln(x² + 4) + (4/5)arctan(x/2) + c.
A quadratic factor in the denominator always takes Bx + C on top. Writing a constant there is the single commonest error on this question type, and it loses every mark after the first line.
When recognition runs out: trig substitution
The two patterns are read-off results, but the same quadratics turn up in shapes the table does not name: √(a² − x²) on its own, or (a² + x²) raised to a power. Those yield to a substitution chosen so a Pythagorean identity collapses the quadratic. Seeing a² − x², set x = a sin θ, because a² − a²sin²θ = a²cos²θ. Seeing a² + x², set x = a tan θ, because a² + a²tan²θ = a²sec²θ. The right substitution is the one that removes the square root or the awkward power.
WORKED EXAMPLE
x = a sin θ on a square root
Evaluate ∫ √(4 − x²) dx from 0 to 2.
The root fits neither standard pattern, so substitute. Set x = 2 sin θ, so dx = 2 cos θ dθ, and the limits x = 0 and x = 2 become θ = 0 and θ = π/2.
√(4 − 4 sin²θ) = 2 cos θ, taking the positive root since cos θ ≥ 0 across [0, π/2]. The integral is ∫ 4 cos²θ dθ = ∫ 2(1 + cos 2θ) dθ.
= [2θ + sin 2θ] from 0 to π/2 = π.
That is a quarter of the circle x² + y² = 4, area πr²/4 with r = 2. The substitution recovered the geometry.
WORKED EXAMPLE
x = a tan θ on a squared quadratic
Find ∫ 1/(1 + x²)² dx from 0 to 1.
Squared, the denominator is beyond the arctan pattern. Set x = tan θ, so dx = sec²θ dθ and 1 + x² = sec²θ.
The integrand becomes sec²θ/sec⁴θ = cos²θ, and the limits x = 0 and x = 1 become θ = 0 and θ = π/4.
∫ cos²θ dθ = ∫ ½(1 + cos 2θ) dθ = [θ/2 + (sin 2θ)/4] from 0 to π/4.
= π/8 + 1/4. The arctan pattern alone had no way to produce that 1/4; the substitution earns it.
With definite limits, convert the limits once and stay in θ to the end. For an indefinite integral the answer has to come back to x: write θ = arcsin(x/a) or arctan(x/a), expand any sin 2θ first, and rebuild sin θ and cos θ from the right triangle that defined the substitution.
ASSESSMENT FOCUS
- Derive arcsin's derivative by implicit differentiation when asked. Quoting it is not a derivation in a 'show that'.
- Identify a before integrating. a² is what sits with the constant, so 9 + x² means a = 3.
- The arctan pattern carries a 1/a factor and the arcsin one does not. Mixing them up is the classic slip.
- Complete the square before reaching for any standard form; a bare x² + bx + c fits nothing.
- Exact answers in π are expected whenever the limits are friendly points of tan or sin.
- When the quadratic is rooted, squared or otherwise beyond the two patterns, substitute: x = a sin θ collapses a² − x², and x = a tan θ collapses a² + x².
CHECK YOURSELF
Evaluate ∫ 1/(1 + x²) dx from 0 to 1, exactly.
Show a hint
arctan at the two limits.
Show the answer
[arctan x] from 0 to 1 = arctan 1 − arctan 0 = π/4 − 0 = π/4.
Differentiate inverses implicitly. sin y = x gives dy/dx = 1/√(1 − x²), with the trig gone.
∫1/√(a² − x²) = arcsin(x/a) and ∫1/(a² + x²) = (1/a) arctan(x/a).
Complete the square, split numerators, and give quadratic factors a Bx + C numerator.
Beyond the patterns, substitute: x = a sin θ collapses a² − x², x = a tan θ collapses a² + x².
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the calculus with inverse trigonometric functions questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Differentiate arcsin x, arccos x and arctan x from scratch via implicit differentiation.
- Integrate 1/√(a² − x²) and 1/(a² + x²) by recognising the patterns.
- Complete the square and use partial fractions to reach the same two patterns.
- Choose and carry out the substitutions x = a sin θ and x = a tan θ for the associated forms.
Open the full revision checklist to see every objective in the course in one place.