Calculus Study Guide: Inverse Trigonometric Functions, Derivatives, and Integrals

Fundamental Properties of Inverse Trigonometric Functions

  • Periodicity and Non-Invertibility of Trigonometric Functions

    • The sine function, sin⁡(x)\sin(x), is defined for all real numbers x∈(−∞,∞)x \in (-\infty, \infty).
    • It is a periodic function that repeats its values infinitely in both directions beyond its central period of 2π2\pi.
    • By virtue of being periodic, the sine function fails the horizontal line test and is not one-to-one.
    • A function is defined as one-to-one if and only if every output value yy corresponds to exactly one input value xx.
    • For sin⁡(x)\sin(x), a single output such as y=1y = 1 occurs at infinitely many inputs xx (e.g., x=π/2,5π/2,−3π/2x = \pi/2, 5\pi/2, -3\pi/2).
    • In foundational algebra, a function must be one-to-one over its domain to be invertible. Therefore, full trigonometric functions are not invertible across their complete domains.

  • Algebraic Precedent for Restricting Domains

    • Restricting the domain of a non-invertible function to create an invertible piece is standard in algebra.
    • The standard quadratic function y=x2y = x^2 is not one-to-one over (−∞,∞)(-\infty, \infty), making it non-invertible as a whole.
    • To define the principal square root function y=xy = \sqrt{x}, the domain of y=x2y = x^2 is restricted to the non-negative real numbers, x≥0x \ge 0 (the right half of the parabola).
    • The function y=x2y = x^2 restricted to x≥0x \ge 0 is one-to-one and has the inverse y=xy = \sqrt{x}.
    • The negative square root, y=−xy = -\sqrt{x}, represents the inverse of the left half of the parabola, restricted to x≤0x \le 0.

  • The Central Cut of Trigonometric Functions

    • To define inverse trigonometric functions, a specific interval where the function is one-to-one must be selected.
    • The standard standard domain choice for y=sin⁡(x)y = \sin(x) is known as the central cut.
    • The central cut for sin⁡(x)\sin(x) is defined on the closed interval [−π/2,π/2][-\pi/2, \pi/2].
    • Quantitative specifics of the central cut:
    • Left endpoint: x=−π/2≈−1.57x = -\pi/2 \approx -1.57, where y=−1y = -1.
    • Point of symmetry: (0,0)(0, 0).
    • Right endpoint: x=π/2≈1.57x = \pi/2 \approx 1.57, where y=1y = 1.
    • Properties of the central cut:
    1. It is one-to-one; every yy-value in [−1,1][-1, 1] occurs exactly once.
    2. It is the largest continuous interval where the function remains one-to-one. Extending beyond π/2\pi/2 or −π/2-\pi/2 causes the graph to double back, violating the one-to-one property.

  • Geometric Reflection and Algebraic Definition of Arcsine

    • Inverse functions are geometric reflections of each other across the line y=xy = x.
    • Under reflection across y=xy = x, input xx-values and output yy-values interchange:
    • For restricted y=sin⁡(x)y = \sin(x): Domain is [−π/2,π/2][-\pi/2, \pi/2] and Range is [−1,1][-1, 1].
    • For y=arcsin⁡(x)y = \arcsin(x) (also written as y=sin⁡−1(x)y = \sin^{-1}(x)): Domain is [−1,1][-1, 1] and Range is [−π/2,π/2][-\pi/2, \pi/2].
    • Formal Definition:     arcsin⁡(x)=y  ⟺  sin⁡(y)=x\arcsin(x) = y \iff \sin(y) = x
    • Cancellation Property:     arcsin⁡(sin⁡(x))=x\arcsin(\sin(x)) = x     This cancellation identity holds if and only if x∈[−π/2,π/2]x \in [-\pi/2, \pi/2].
    • If an input lies outside the restricted domain (e.g., x=3π/2x = 3\pi/2), arcsin⁡(sin⁡(3π/2))≠3π/2\arcsin(\sin(3\pi/2)) \neq 3\pi/2 because the output of arcsin⁡(x)\arcsin(x) must strictly fall within [−π/2,π/2][-\pi/2, \pi/2].

The Derivative of an Inverse Function Theorem

  • Theorem Statement (Calculus I, Section 5.3)

    • Let f(x)f(x) and g(x)g(x) be inverse functions such that f(g(x))=xf(g(x)) = x and g(f(x))=xg(f(x)) = x.
    • If f(x)f(x) is differentiable and f′(g(x))≠0f'(g(x)) \neq 0, then g(x)g(x) is differentiable, and its derivative is given by:     g′(x)=1f′(g(x))g'(x) = \frac{1}{f'(g(x))}
    • Alternative Notation:     (f−1)′(x)=1f′(f−1(x))(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}

  • Formal Proof via the Chain Rule

    • Start with the composition identity of inverse functions:     f(g(x))=xf(g(x)) = x
    • Differentiate both sides of the equation with respect to xx using the Chain Rule:     ddx[f(g(x))]=ddx[x]\frac{d}{dx}[f(g(x))] = \frac{d}{dx}[x]f′(g(x))⋅g′(x)=1f'(g(x)) \cdot g'(x) = 1
    • Solve for g′(x)g'(x) by dividing both sides by f′(g(x))f'(g(x)):     g′(x)=1f′(g(x))g'(x) = \frac{1}{f'(g(x))}

  • Standard Derivatives of Prerequisite Trigonometric Functions

    • ddx[sin⁡(x)]=cos⁡(x)\frac{d}{dx}[\sin(x)] = \cos(x)
    • ddx[cos⁡(x)]=−sin⁡(x)\frac{d}{dx}[\cos(x)] = -\sin(x)
    • ddx[tan⁡(x)]=sec⁡2(x)\frac{d}{dx}[\tan(x)] = \sec^2(x)
    • ddx[sec⁡(x)]=sec⁡(x)tan⁡(x)\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)
    • ddx[csc⁡(x)]=−csc⁡(x)cot⁡(x)\frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x)
    • ddx[cot⁡(x)]=−csc⁡2(x)\frac{d}{dx}[\cot(x)] = -\csc^2(x)

Derivation of Derivatives for Inverse Trigonometric Functions

  • Derivation of the Derivative of arcsin⁡(x)\arcsin(x)

    • Define g(x)=arcsin⁡(x)g(x) = \arcsin(x) and its inverse function f(x)=sin⁡(x)f(x) = \sin(x).
    • Compute the derivative of f(x)f(x): f′(x)=cos⁡(x)f'(x) = \cos(x).
    • Apply the Inverse Function Differentiation Theorem:     g′(x)=ddx[arcsin⁡(x)]=1cos⁡(arcsin⁡(x))g'(x) = \frac{d}{dx}[\arcsin(x)] = \frac{1}{\cos(\arcsin(x))}

  • Geometric Simplification Using a Reference Triangle

    • Every inverse trigonometric output represents an angle: let θ=arcsin⁡(x)\theta = \arcsin(x), which implies sin⁡(θ)=x=x1\sin(\theta) = x = \frac{x}{1}.
    • In a right-angled triangle with angle θ\theta:
    • Opposite side = xx
    • Hypotenuse = 11
    • By the Pythagorean Theorem (a2+b2=c2a^2 + b^2 = c^2), find the adjacent side:     Adjacent2+x2=12\text{Adjacent}^2 + x^2 = 1^2Adjacent=1−x2\text{Adjacent} = \sqrt{1 - x^2}
    • Evaluate cos⁡(θ)=AdjacentHypotenuse\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}:     cos⁡(arcsin⁡(x))=1−x21=1−x2\cos(\arcsin(x)) = \frac{\sqrt{1 - x^2}}{1} = \sqrt{1 - x^2}
    • Substitute this back into the derivative formula:     ddx[arcsin⁡(x)]=11−x2\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1 - x^2}}

Explicit Differentiation Rules for Inverse Trigonometric Functions

  • General Differentiation Rules with Chain Rule (u=u(x)u = u(x))

    • Let uu be a differentiable function of xx. Applying the Chain Rule yields the general forms:
    • Arcsine:     ddx[arcsin⁡(u)]=u′1−u2\frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1 - u^2}}
    • Arctangent:     ddx[arctan⁡(u)]=u′1+u2\frac{d}{dx}[\arctan(u)] = \frac{u'}{1 + u^2}
    • Arcsecant:     \frac{d}{dx}[\arcsec(u)] = \frac{u'}{|u|\sqrt{u^2 - 1}}

  • Derivation and Significance of the Absolute Value in \arcsec(u)

    • During the algebraic derivation of \frac{d}{dx}[\arcsec(u)], the term u4−u2\sqrt{u^4 - u^2} appears in the denominator.
    • Factoring out u2u^2 yields u2(u2−1)=u2u2−1\sqrt{u^2(u^2 - 1)} = \sqrt{u^2}\sqrt{u^2 - 1}.
    • By algebraic definition, u2=∣u∣\sqrt{u^2} = |u|.
    • Example: If u=5u = 5, then 52=5\sqrt{5^2} = 5. If u=−5u = -5, then (−5)2=25=5=∣−5∣\sqrt{(-5)^2} = \sqrt{25} = 5 = |-5|. The absolute value ensures the expression remains positive regardless of the sign of uu.

  • Cofunction Differentiation Rules

    • The derivatives of the cofunctions (Arccosine, Arccotangent, Arccosecant) match their corresponding non-co functions exactly, with an added negative sign in the numerator:
    • Arccosine:     ddx[arccos⁡(u)]=−u′1−u2\frac{d}{dx}[\arccos(u)] = -\frac{u'}{\sqrt{1 - u^2}}
    • Arccotangent:     ddx[arccot(u)]=−u′1+u2\frac{d}{dx}[\text{arccot}(u)] = -\frac{u'}{1 + u^2}
    • Arccosecant:     ddx[arccsc(u)]=−u′∣u∣u2−1\frac{d}{dx}[\text{arccsc}(u)] = -\frac{u'}{|u|\sqrt{u^2 - 1}}

Worked Differentiation Examples and Applications

  • Example 1: Differentiating arcsin⁡(ln⁡(x))\arcsin(\ln(x))

    • Let u=ln⁡(x)u = \ln(x), so u′=1xu' = \frac{1}{x}.
    • Apply the formula u′1−u2\frac{u'}{\sqrt{1 - u^2}}:     ddx[arcsin⁡(ln⁡(x))]=1x1−(ln⁡(x))2\frac{d}{dx}[\arcsin(\ln(x))] = \frac{\frac{1}{x}}{\sqrt{1 - (\ln(x))^2}}
    • Simplify the complex fraction:     ddx[arcsin⁡(ln⁡(x))]=1x1−(ln⁡(x))2\frac{d}{dx}[\arcsin(\ln(x))] = \frac{1}{x\sqrt{1 - (\ln(x))^2}}

  • Example 2: Differentiating arccot(2x)\text{arccot}(2^x)

    • Let u=2xu = 2^x, so u′=2xln⁡(2)u' = 2^x \ln(2).
    • Apply the formula −u′1+u2-\frac{u'}{1 + u^2}:     ddx[arccot(2x)]=−2xln⁡(2)1+(2x)2\frac{d}{dx}[\text{arccot}(2^x)] = -\frac{2^x \ln(2)}{1 + (2^x)^2}
    • Apply exponent rules (ab)c=abc(a^b)^c = a^{bc} to simplify denominator:     ddx[arccot(2x)]=−2xln⁡(2)1+22x\frac{d}{dx}[\text{arccot}(2^x)] = -\frac{2^x \ln(2)}{1 + 2^{2x}}

  • Example 3: Differentiating arcsin⁡(x)\arcsin(\sqrt{x})

    • Let u=x=x1/2u = \sqrt{x} = x^{1/2}, so u′=12xu' = \frac{1}{2\sqrt{x}}.
    • Apply the formula:     ddx[arcsin⁡(x)]=12x1−(x)2=12x1−x\frac{d}{dx}[\arcsin(\sqrt{x})] = \frac{\frac{1}{2\sqrt{x}}}{\sqrt{1 - (\sqrt{x})^2}} = \frac{1}{2\sqrt{x}\sqrt{1 - x}}
    • Combine radicals in the denominator:     ddx[arcsin⁡(x)]=12x(1−x)=12x−x2\frac{d}{dx}[\arcsin(\sqrt{x})] = \frac{1}{2\sqrt{x(1 - x)}} = \frac{1}{2\sqrt{x - x^2}}

  • Example 4: Differentiating xarctan⁡(x)x \arctan(x)

    • Apply the Product Rule ddx[uv]=u′v+uv′\frac{d}{dx}[uv] = u'v + uv' where u=xu = x and v=arctan⁡(x)v = \arctan(x):     ddx[xarctan⁡(x)]=(1)⋅arctan⁡(x)+x⋅(11+x2)\frac{d}{dx}[x \arctan(x)] = (1)\cdot\arctan(x) + x\cdot\left(\frac{1}{1 + x^2}\right)ddx[xarctan⁡(x)]=arctan⁡(x)+x1+x2\frac{d}{dx}[x \arctan(x)] = \arctan(x) + \frac{x}{1 + x^2}

  • Example 5: Comprehensive Problem (Homework #55)

    • Problem Statement: Differentiate y=8arcsin⁡(x4)−x16−x22y = 8\arcsin\left(\frac{x}{4}\right) - \frac{x\sqrt{16 - x^2}}{2}.
    • Step 1: Differentiate the first term, 8arcsin⁡(x4)8\arcsin\left(\frac{x}{4}\right):     ddx[8arcsin⁡(x4)]=8⋅141−(x4)2=21−x216=216−x216=216−x24=816−x2\frac{d}{dx}\left[8\arcsin\left(\frac{x}{4}\right)\right] = 8 \cdot \frac{\frac{1}{4}}{\sqrt{1 - \left(\frac{x}{4}\right)^2}} = \frac{2}{\sqrt{1 - \frac{x^2}{16}}} = \frac{2}{\sqrt{\frac{16 - x^2}{16}}} = \frac{2}{\frac{\sqrt{16 - x^2}}{4}} = \frac{8}{\sqrt{16 - x^2}}
    • Step 2: Differentiate the second term, −12(x16−x2)-\frac{1}{2}\left(x\sqrt{16 - x^2}\right), using the Product Rule:     ddx[x16−x2]=(1)16−x2+x(−2x216−x2)=16−x2−x216−x2\frac{d}{dx}\left[x\sqrt{16 - x^2}\right] = (1)\sqrt{16 - x^2} + x\left(\frac{-2x}{2\sqrt{16 - x^2}}\right) = \sqrt{16 - x^2} - \frac{x^2}{\sqrt{16 - x^2}}
    • Step 3: Combine inside terms over common denominator 16−x2\sqrt{16 - x^2}:     16−x2−x216−x2=(16−x2)2−x216−x2=16−x2−x216−x2=16−2x216−x2\sqrt{16 - x^2} - \frac{x^2}{\sqrt{16 - x^2}} = \frac{(\sqrt{16 - x^2})^2 - x^2}{\sqrt{16 - x^2}} = \frac{16 - x^2 - x^2}{\sqrt{16 - x^2}} = \frac{16 - 2x^2}{\sqrt{16 - x^2}}
    • Step 4: Multiply by the front factor −12-\frac{1}{2}:     −12⋅16−2x216−x2=−8+x216−x2-\frac{1}{2} \cdot \frac{16 - 2x^2}{\sqrt{16 - x^2}} = \frac{-8 + x^2}{\sqrt{16 - x^2}}
    • Step 5: Combine steps 1 and 4:     y′=816−x2+−8+x216−x2=8−8+x216−x2=x216−x2y' = \frac{8}{\sqrt{16 - x^2}} + \frac{-8 + x^2}{\sqrt{16 - x^2}} = \frac{8 - 8 + x^2}{\sqrt{16 - x^2}} = \frac{x^2}{\sqrt{16 - x^2}}

Derivatives Applied to Tangent Line Equations

  • Homework Problem #61: Finding a Tangent Line Equation

    • Problem Statement: Find the equation of the line tangent to y=4xarccos⁡(x−1)y = 4x \arccos(x - 1) at the specific point (1,2π)(1, 2\pi).

    • Step 1: Compute the derivative y′y' using the Product and Chain Rules:     y′=ddx[4x]⋅arccos⁡(x−1)+4x⋅ddx[arccos⁡(x−1)]y' = \frac{d}{dx}[4x]\cdot\arccos(x - 1) + 4x\cdot\frac{d}{dx}[\arccos(x - 1)]y′=4arccos⁡(x−1)+4x(−11−(x−1)2)y' = 4\arccos(x - 1) + 4x\left(-\frac{1}{\sqrt{1 - (x - 1)^2}}\right)y′=4arccos⁡(x−1)−4x1−(x−1)2y' = 4\arccos(x - 1) - \frac{4x}{\sqrt{1 - (x - 1)^2}}

    • Step 2: Evaluate the derivative at x=1x = 1 to determine slope mm:     m=y′(1)=4arccos⁡(1−1)−4(1)1−(1−1)2m = y'(1) = 4\arccos(1 - 1) - \frac{4(1)}{\sqrt{1 - (1 - 1)^2}}m=4arccos⁡(0)−41−0m = 4\arccos(0) - \frac{4}{\sqrt{1 - 0}}

    • Note: arccos⁡(0)=π/2\arccos(0) = \pi/2 radians (90∘90^\circ is not usable in calculus equations as degrees cannot combine algebraically with real numbers).     m=4(π2)−4=2π−4m = 4\left(\frac{\pi}{2}\right) - 4 = 2\pi - 4

    • Step 3: Apply point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point (1,2π)(1, 2\pi) and slope m=2π−4m = 2\pi - 4:     y−2π=(2π−4)(x−1)y - 2\pi = (2\pi - 4)(x - 1)y−2π=(2π−4)x−(2π−4)y - 2\pi = (2\pi - 4)x - (2\pi - 4)y−2π=(2π−4)x−2π+4y - 2\pi = (2\pi - 4)x - 2\pi + 4y=(2π−4)x+4y = (2\pi - 4)x + 4

Integration Involving Inverse Trigonometric Functions

  • Fundamental Integration Formulas (Section 5.8)

    • Reversing differentiation formulas yields core indefinite integrals. Let uu be a differentiable function of xx, and let a>0a > 0 be a positive real constant:

    1. Arcsine Form:      ∫dua2−u2=arcsin⁡(ua)+C\int \frac{du}{\sqrt{a^2 - u^2}} = \arcsin\left(\frac{u}{a}\right) + C

    2. Arctangent Form:      ∫dua2+u2=1aarctan⁡(ua)+C\int \frac{du}{a^2 + u^2} = \frac{1}{a} \arctan\left(\frac{u}{a}\right) + C

    3. Arcsecant Form:      \int \frac{du}{|u|\sqrt{u^2 - a^2}} = \frac{1}{a} \arcsec\left(\frac{|u|}{a}\right) + C

  • Redundancy of Cofunction Integrals

    • Separate integration formulas for cofunctions are unnecessary because negative signs factor out of integrals directly:     ∫−dua2+u2=−∫dua2+u2=−1aarctan⁡(ua)+C\int -\frac{du}{a^2 + u^2} = -\int \frac{du}{a^2 + u^2} = -\frac{1}{a}\arctan\left(\frac{u}{a}\right) + C

Integration Techniques and Substitution Examples

  • Example 1: Evaluating ∫x9+x4 dx\int \frac{x}{9 + x^4} \, dx

    • Pattern Recognition: The denominator contains no square root and fits the sum of squares form a2+u2a^2 + u^2, pointing toward the Arctangent integral formula.
    • Rewrite the integrand terms:     9+x4=32+(x2)2  ⟹  a=3,u=x29 + x^4 = 3^2 + (x^2)^2 \implies a = 3, \quad u = x^2
    • Compute differential dudu:     du=2x dx  ⟹  x dx=du2du = 2x \, dx \implies x \, dx = \frac{du}{2}
    • Adjust constants inside and outside the integral:     ∫x9+x4 dx=12∫2x dx32+(x2)2=12∫du32+u2\int \frac{x}{9 + x^4} \, dx = \frac{1}{2} \int \frac{2x \, dx}{3^2 + (x^2)^2} = \frac{1}{2} \int \frac{du}{3^2 + u^2}
    • Apply the arctangent integration formula:     12[13arctan⁡(u3)]+C=16arctan⁡(u3)+C\frac{1}{2} \left[ \frac{1}{3} \arctan\left(\frac{u}{3}\right) \right] + C = \frac{1}{6} \arctan\left(\frac{u}{3}\right) + C
    • Substitute back u=x2u = x^2:     ∫x9+x4 dx=16arctan⁡(x23)+C\int \frac{x}{9 + x^4} \, dx = \frac{1}{6} \arctan\left(\frac{x^2}{3}\right) + C

  • Example 2: Setting up ∫5x16−52x dx\int \frac{5^x}{\sqrt{16 - 5^{2x}}} \, dx

    • Pattern Recognition: The presence of a square root in the denominator of the form a2−u2\sqrt{a^2 - u^2} indicates an Arcsine integration pattern.
    • Identify components:     16−52x=42−(5x)2  ⟹  a=4,u=5x16 - 5^{2x} = 4^2 - (5^x)^2 \implies a = 4, \quad u = 5^x
    • Compute differential dudu:     du=5xln⁡(5) dxdu = 5^x \ln(5) \, dx