Tags: arcsin, arccos, arctan, arcsec, inverse trig derivatives, inverse trig integrals, completing the square, hyperbolic functions, sinh, cosh, tanh, coth, inverse hyperbolic, Calculus, Texas A&M, Chapter 5
Sections 5.7–5.8 cover the derivatives and integrals of the six inverse trigonometric functions, with special emphasis on the three integral forms (arcsin, arctan, arcsec) and how to spot them by completing the square or rewriting the integrand. Section 5.9 introduces hyperbolic functions (sinh, cosh, tanh, etc.), their derivatives, integrals, and identities, plus inverse hyperbolic function derivatives.
Inverse sine, arcsin(x)
The angle whose sine is x. Domain: [−1, 1]. Range: [−π/2, π/2].
Inverse tangent, arctan(x)
The angle whose tangent is x. Domain: all reals. Range: (−π/2, π/2).
Inverse secant, arcsec(x)
The angle whose secant is x. Domain: |x| ≥ 1. Range: [0, π], excluding π/2.
Completing the square
An algebraic technique that rewrites ax² + bx + c in the form a(x − h)² + k. Essential for converting integrands into recognisable arcsin or arctan forms.
Hyperbolic sine, sinh(x)
Defined as (e^x − e^(−x))/2. Analogous to sin(x) but built from exponentials.
Hyperbolic cosine, cosh(x)
Defined as (e^x + e^(−x))/2. Always ≥ 1. Analogous to cos(x).
Hyperbolic tangent, tanh(x)
sinh(x)/cosh(x) = (e^x − e^(−x))/(e^x + e^(−x)). Outputs lie in (−1, 1).
Function | Derivative |
|---|---|
arcsin(u) | u' / √(1 − u²) |
arccos(u) | −u' / √(1 − u²) |
arctan(u) | u' / (1 + u²) |
arccot(u) | −u' / (1 + u²) |
arcsec(u) | u' / (|u|√(u² − 1)) |
arccsc(u) | −u' / (|u|√(u² − 1)) |
Converting to algebraic form
For tan(arcsec(x/5)):
Let θ = arcsec(x/5), so sec θ = x/5
Draw a right triangle: hypotenuse = x, adjacent = 5, opposite = √(x² − 25)
tan θ = √(x² − 25)/5
Solving inverse trig equations
For arcsin(4x − π) = 1/2:
Apply sin to both sides: 4x − π = sin(1/2)
x = (π + sin(1/2))/4
Derivative of f(x) = 4 arcsin(x − 2)
f'(x) = 4/√(1 − (x − 2)²) = 4/√(4x − x² − 3)
Derivative of f(x) = arcsec(8x)
f'(x) = 8/(|8x|√(64x² − 1)) = 1/(|x|√(64x² − 1))
Combined derivative (arctan + rational)
For y = arctan(x/6) + (2x − 3)/(6(x² + 1)):
The arctan part gives (1/6) · 1/(1 + (x/6)²)
The rational part requires the quotient rule
After simplification the two parts combine neatly
dy/dx = (1/6) · [1/(1 + x²/36) + (2 + 6x − 2x²)/(x² + 1)²]
Tangent line to y = arcsin(2x)
At the point (1/(2√2), π/4):
dy/dx = 2/√(1 − 4x²)
At x = 1/(2√2): slope = 2/√(1 − 1/2) = 2/√(1/2) = 2√2
Tangent line: y = 2√2 · x + π/4 − 1
The three key integral patterns:
∫ 1/√(a² − u²) du = arcsin(u/a) + C
∫ 1/(a² + u²) du = (1/a) arctan(u/a) + C
∫ 1/(u√(u² − a²)) du = (1/a) arcsec(|u|/a) + C
Recognising the arcsin pattern
∫ 1/√(81 − 121x²) dx:
Here a² = 81 (a = 9) and u = 11x, du = 11 dx
(1/11) arcsin(11x/9) + C
Recognising the arcsec pattern
∫ 1/(x√(196x² − 81)) dx:
Rewrite as ∫ 1/(x · √((14x)² − 81)) dx
a = 9, u = 14x
(1/9) arcsec(|14x|/9) + C
Recognising the arctan pattern
∫ 1/(4 + (x − 4)²) dx:
a² = 4 (a = 2), u = x − 4
(1/2) arctan((x − 4)/2) + C
Substitution to create arctan
∫ t/(t⁴ + 36) dt:
Let u = t², du = 2t dt
(1/2) ∫ 1/(u² + 36) du = (1/2) · (1/6) arctan(u/6) + C
= (1/12) arctan(t²/6) + C
Splitting integrals (ln + arctan)
∫ (2x − 4)/(x² + 10x + 50) dx:
Complete the square: x² + 10x + 50 = (x + 5)² + 25
Split the numerator to match the derivative of the denominator and a constant:
2x − 4 = 2(x + 5) − 14 (since derivative of x² + 10x + 50 is 2x + 10)
∫ 2(x + 5)/((x + 5)² + 25) dx − 14 ∫ 1/((x + 5)² + 25) dx
= ln|x² + 10x + 50| − (14/5) arctan((x + 5)/5) + C
Another split: ∫ (x − 25)/(x² + 1) dx
Split: x/(x² + 1) − 25/(x² + 1)
= (1/2) ln(x² + 1) − 25 arctan(x) + C
Completing the square for arcsin
∫ 1/√(−x² − 12x) dx:
−x² − 12x = −(x² + 12x) = −((x + 6)² − 36) = 36 − (x + 6)²
∫ 1/√(36 − (x + 6)²) dx = arcsin((x + 6)/6) + C
Initial value problem with arcsin
dy/dx = 3/√(81 − x²), y(0) = π:
Integrate: y = 3 arcsin(x/9) + C
Use y(0) = π: π = 3 arcsin(0) + C, so C = π
y = 3 arcsin(x/9) + π
Definitions
sinh(x) = (e^x − e^(−x))/2
cosh(x) = (e^x + e^(−x))/2
tanh(x) = sinh(x)/cosh(x)
coth(x) = cosh(x)/sinh(x)
sech(x) = 1/cosh(x)
csch(x) = 1/sinh(x)
Key identity
cosh²(x) − sinh²(x) = 1 (analogous to cos² + sin² = 1, note the sign difference)
Evaluating from definitions
sinh(ln 6) = (e^(ln 6) − e^(−ln 6))/2 = (6 − 1/6)/2 = 35/12
cosh(ln 10) = (e^(ln 10) + e^(−ln 10))/2 = (10 + 1/10)/2 = 101/20
Derivatives
Function | Derivative |
|---|---|
sinh(u) | cosh(u) · u' |
cosh(u) | sinh(u) · u' |
tanh(u) | sech²(u) · u' |
coth(u) | −csch²(u) · u' |
sech(u) | −sech(u) tanh(u) · u' |
csch(u) | −csch(u) coth(u) · u' |
Derivative of h(x) = (5/4) sinh(2x) − 5x/2
h'(x) = (5/4) · cosh(2x) · 2 − 5/2 = (5/2) cosh(2x) − 5/2
Using cosh²(u) − sinh²(u) = 1 and the double angle: this simplifies to 5 sinh²(x)
Derivative of y = coth(4x)
dy/dx = −csch²(4x) · 4 = −4 csch²(4x)
Hyperbolic integrals
∫ sinh(u) du = cosh(u) + C
∫ cosh(u) du = sinh(u) + C
∫ sech²(u) du = tanh(u) + C
∫ tanh(u) du = ln(cosh(u)) + C
Integral of sinh(2 − 5x)
Let u = 2 − 5x, du = −5 dx
−(1/5) cosh(2 − 5x) + C
Definite integral of 5 tanh(x) from 0 to ln 7
∫ tanh(x) dx = ln(cosh(x))
5[ln(cosh(ln 7)) − ln(cosh(0))]
cosh(ln 7) = (7 + 1/7)/2 = 25/7, and cosh(0) = 1
= 5 ln(25/7)
Inverse hyperbolic derivative
For y = 7 sinh⁻¹(tan 2x):
d/dx [sinh⁻¹(u)] = u'/√(1 + u²)
u = tan(2x), u' = 2 sec²(2x)
dy/dx = 7 · 2 sec²(2x)/√(1 + tan²(2x))
Since 1 + tan² = sec², √(sec²(2x)) = |sec(2x)|
dy/dx = 14 sec²(2x)/|sec(2x)| = 14|sec(2x)|
Three integral forms to memorise (inverse trig):
∫ 1/√(a² − u²) du = arcsin(u/a) + C
∫ 1/(a² + u²) du = (1/a) arctan(u/a) + C
∫ 1/(u√(u² − a²)) du = (1/a) arcsec(|u|/a) + C
Hyperbolic identities:
cosh²(x) − sinh²(x) = 1
sinh(2x) = 2 sinh(x) cosh(x)
cosh(2x) = cosh²(x) + sinh²(x)
Inverse hyperbolic derivatives:
d/dx [sinh⁻¹(u)] = u'/√(1 + u²)
d/dx [cosh⁻¹(u)] = u'/√(u² − 1)
d/dx [tanh⁻¹(u)] = u'/(1 − u²)
⚠️ The three inverse trig integral patterns are some of the most tested formulas in this chapter. You need to recognise each one on sight and know which substitution to use.
⚠️ Completing the square is the critical algebraic skill for this section. If the integrand has a quadratic under a square root or in a denominator, complete the square first to reveal the arcsin or arctan form.
⚠️ When an integral has a linear term in the numerator over a quadratic (like (2x − 4)/(x² + 10x + 50)), split it into a part that matches the derivative of the denominator (giving ln) and a constant part (giving arctan).
⚠️ For hyperbolic function evaluations, use the exponential definitions directly. sinh(ln a) = (a − 1/a)/2 and cosh(ln a) = (a + 1/a)/2 are handy shortcuts.
⚠️ The derivative of coth is −csch², and the derivative of tanh is sech². Note the sign differences from the trig analogues.
⚠️ For ∫ t/(t⁴ + 36) dt, the substitution u = t² converts t⁴ into u², reducing it to an arctan integral. Look for this pattern when you see an odd power of t in the numerator.
Q: What is tan(arcsec(x/5)) in algebraic form?
A: √(x² − 25)/5. Draw a right triangle with hypotenuse x, adjacent 5, and compute the opposite side.
Q: Find ∫ 1/√(81 − 121x²) dx.
A: (1/11) arcsin(11x/9) + C. Here a = 9 and u = 11x.
Q: Find ∫ 1/(4 + (x − 4)²) dx.
A: (1/2) arctan((x − 4)/2) + C. The form matches the arctan pattern with a = 2, u = x − 4.
Q: How do you handle ∫ (2x − 4)/(x² + 10x + 50) dx?
A: Complete the square in the denominator to get (x + 5)² + 25. Split the numerator as 2(x + 5) − 14. The first part integrates to ln, the second to arctan.
Q: What is sinh(ln 6)?
A: 35/12. Use the definition: (6 − 1/6)/2 = (36 − 1)/12.
Q: Find the derivative of y = coth(4x).
A: dy/dx = −4 csch²(4x).
Q: Evaluate ∫ from 0 to ln 7 of 5 tanh(x) dx.
A: 5 ln(25/7). The antiderivative of tanh(x) is ln(cosh(x)).
inverse trig functions, arcsin, arccos, arctan, arcsec, arccsc, inverse trig derivatives, inverse trig integrals, completing the square integration, arctan integral pattern, arcsin integral pattern, arcsec integral pattern, hyperbolic functions, sinh, cosh, tanh, coth, sech, csch, hyperbolic identities, inverse hyperbolic derivatives, sinh inverse, completing the square, splitting integrals, Calculus Chapter 5, Section 5.7, Section 5.8, Section 5.9, Texas A&M Calculus