Unit 1 Calculus 2

∫ tanx dx

ln|secx|

∫ cotx dx

ln|sinx|

∫ secx dx

ln|secx + tanx|

∫ cscx dx

-ln|cscx + cotx|

Domain/range restriction for sinθ = x / sin^-1x = θ

interval: -π/2 ≤ θ ≤ π/2

Unit Circle: Q4 and Q1

Domain/range restriction for tanθ = x / tan^-1x = θ

interval: -π/2 < θ < π/2

Unit Circle: Q4 and Q1

Domain/range restriction for cosθ = x / cos^-1x = θ

interval: 0 ≤ θ ≤ π

Unit Circle: Q1 and Q2

Domain/range restriction for secθ = x / sec^-1x = θ

interval: θ ∈ [0,π/2) U [π,3π/2)

Unit Circle: Q1 and Q3

How to find the exact value of an inverse function

1. Let f^-1(a) = θ, so f(θ) = a

2. Locate θ on the Unit Circle

How to simplify g(f^-1(a))

1. Let f^-1(a) = θ

2. Put f(θ) = a on a right triangle

3. Find g(θ)

   * A leg can be negative depending on the quadrant of the Unit Circle, but the hypotenuse is always positive.

Graph of tan^-1(x)

• tan^-1(0) = 0

• tan^-1(+∞) = π/2

• tan^-1(-∞) = -π/2

d/dx sin^-1x

d/dx cos^-1x

d/dx tan^-1x

d/dx cot^-1x

d/dx sec^-1x

d/dx csc^-1x

How to derive an inverse function by implicit differentiation

1. Let f^-1(x) = y, so f(y) = x

2. d/dx [f(y) = x]

3. Solve for dy/dx

4. Re-express dy/dx using the Pythagorean identities

5. Substitute x for f(y)

6. d/dx [f^-1(y) = x]

7. Replace dy/dx with terms of x

sin^-1x

tan^-1x

sec^-1x

Integration by parts formula

indefinite integral: ∫ u dv = uv − ∫ v du

Tips

• What’s easy to derive is u. What’s easy to integrate is dv.

• Choose the x with the highest power to be u.

• Sometimes substitution is needed to set up ∫ u dv.

• If you end up with the original integral, combine and isolate it.

∫ sin^m(x)cosⁿ(x) dx

n ≥ 1 and odd

1. Factor out cosx

2. Substitute cos²x = 1 − sin²x

3. u = sinx, du = cosx dx (replaces the factored cosx)

∫ sin^m(x)cosⁿ(x) dx

m ≥ 1 and odd

1. Factor out sinx

2. Substitute sin²x = 1 − cos²x

3. u = cosx, du = -sinx dx (replaces the factored sinx)

∫ sin^m(x)cosⁿ(x) dx

n and m are even

Substitute:

sin²x = ½ (1 − cos2x)

cos²x = ½ (1 + cos2x)

∫ tan^m(x)secⁿ(x) dx

n ≥ 2 and even

1. Factor out sec²x

2. Substitute sec²x = tan²x + 1

3. u = tanx, du = sec²x dx (replaces the factored sec²x)

∫ tan^m(x)secⁿ(x) dx

n, m ≥ 1 and m is odd

1. Factor out secxtanx

2. Substitute tan²x = sec²x − 1

3. u = secx, du = secxtanx dx (replaces the factored secxtanx)

1. Substitute x = asinθ and

   dx = acosθ dθ

   √(a² − (asinθ)²)

   √(a² − a²sin²θ)

2. Factor out a²

   √(a²(1 − sin²θ))

   a√(1 − sin²θ)

3. Substitute 1 − sin²θ = cos²θ

   a√cos²θ

   acosθ

4. Convert from terms of θ to x using a right triangle, knowing that sinθ = x/a

1. Substitute x = atanθ and

   dx = asec²θ dθ

   √((atanθ)² + a²)

   √(a²tan²θ + a²)

2. Factor out a²

   √(a²(tan²θ + 1))

   a√(tan²θ + 1)

3. Substitute tan²θ + 1 = sec²θ

   a√sec²θ

   asecθ

4. Convert from terms of θ to x using a right triangle, knowing that tanθ = x/a

1. Substitute x = asecθ and

   dx = secθtanθ dθ

   √((asecθ)² − a²)

   √(a²sec²θ − a²)

2. Factor out a²

   √(a²(sec²θ − 1))

   a√(sec²θ − 1)

3. Substitute sec²θ − 1 = tan²θ

   a√tan²θ

   atanθ

4. Convert from terms of θ to x using a right triangle, knowing that secθ = x/a

Complete the square formula for

x² + bx + c

Double angle formula

sin(2x) = 2sinxcosx

numerator degree ≥ denominator degree

Use long division to simplify the integrand.

numerator degree < denominator degree

1. Factor the denominator, if needed.

2. Decompose the integrand into partial fractions.

3. Combine fractions using the same common denominator as the original.

4. Set the new numerator equal to the original.

5. Solve for the unknown constants by testing eliminatory x-values and creating equations from coefficients.

The denominator has distinct linear factors:

(a₁x + b₁) (a₂x + b₂) … (a_nx + b_n)

The denominator has a repeated linear factor:

(ax + b)ⁿ

The denominator has a distinct irreducible factor:

ax² + bx + c

The denominator has a repeated irreducible factor:

(ax² + bx + c)ⁿ

How to rationalize a function with a square root in the denominator using u-substitution

1. Let u equal the square root.

   ex. u = √x

2. Find u² to cancel out the radical.

   u² = (√x)²

   u² = x

3. Derive u².

   2u du = 1dx

4. Substitute.

Integral with respect to x for the area between two curves

Integral with respect to y for the area between two curves

Integral for the volume of a solid...

• without a hole

• revolved around a horizontal axis (y = a)

Disk method with respect to x

Integral for the volume of a solid…

• without a hole

• revolved around a vertical axis (x = a)

Disk method with respect to y

Integral for the volume of a solid…

• with a hole

• revolved around a horizontal axis (y = a)

Washer method with respect to x

* Each radius is upper − lower

Integral for the volume of a solid…

• with a hole

• revolved around a vertical axis (x = a)

Washer method with respect to y

* Each radius is right − left

L’Hôpital’s rule

Conditions: f(a)/g(a) = 0/0 or ∞/∞ (indeterminate form, NOT 1)

Tips

• Can be performed as many times as needed and with higher-order derivatives

• Create one fraction by taking reciprocals and substituting trig identities.

• The limit of a constant is that very constant.

How to break out a radicand

Limit chain rule