Maths › Further calculus › Calculus with inverse trigonometric functions
Calculus with inverse trigonometric functions
Differentiating arcsin and arctan produces algebraic fractions, and read backwards that is a gift: 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.
- Evaluate exact definite integrals that land on π.
WHAT YOU PROBABLY THINK
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: cos y (dy/dx) = 1. Since cos y = √(1 − sin²y) = √(1 − x²), the trig evaporates, which the opening claim never saw coming:
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 → 0: 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 to the exam's two workhorses: ∫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: 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: arctan carries it in.
YOUR TURN
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.
THE EXAM BIT
- Derive arcsin's derivative by implicit differentiation when asked; quoting it earns nothing in a 'show that'.
- Identify a before integrating: a² is what sits with the constant, so 9 + x² means a = 3, not 9.
- The arctan pattern carries a 1/a factor; the arcsin one does not. Mixing them up is the classic slip.
- Exact answers in π are expected whenever the limits are friendly points of tan or sin.
CHECK YOURSELF
Evaluate ∫ 1/(1 + x²) dx from 0 to 1, exactly.
Show a hint
arctan at the two limits.
Show the answer
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Differentiate inverses implicitly: sin y = x gives dy/dx = 1/√(1 − x²), trig gone.
∫1/√(a² − x²) = arcsin(x/a); ∫1/(a² + x²) = (1/a) arctan(x/a). Spot a, write it down.
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.
CHECK YOUR PROGRESS
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- 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.
- Evaluate exact definite integrals that land on π.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.