Calculus 2

Integration Techniques

U-substitution, integration by parts (LIATE & tabular), trig integrals, trig substitution, partial fractions, and improper integrals — every method you need for Calculus 2.

Common Antiderivatives Reference Table

Always check whether the integral matches a basic form before applying a technique. These are the antiderivatives every Calculus 2 student should know cold.

IntegralAntiderivative
∫xⁿ dxxⁿ⁺¹/(n+1) + C (n ≠ −1)
∫1/x dxln|x| + C
∫eˣ dxeˣ + C
∫aˣ dxaˣ / ln a + C
∫sin x dx−cos x + C
∫cos x dxsin x + C
∫sec²x dxtan x + C
∫csc²x dx−cot x + C
∫sec x tan x dxsec x + C
∫csc x cot x dx−csc x + C
∫tan x dxln|sec x| + C
∫sec x dxln|sec x + tan x| + C
∫1/√(1−x²) dxarcsin x + C
∫1/(1+x²) dxarctan x + C
∫1/(x√(x²−1)) dxarcsec|x| + C

U-Substitution (Review)

U-substitution is the chain rule in reverse. It is the first technique to try on any composite integral.

1

Identify u

Choose u = g(x) to be the inner function of a composition. Look for a function whose derivative also appears (or is a constant multiple of something) in the integrand.

2

Compute du

Differentiate: du = g'(x) dx. Solve for dx = du / g'(x) and substitute. If a constant factor is missing, adjust by multiplying and dividing.

3

Rewrite entirely in u

Every x and dx must be replaced. If any x remains, u was not chosen well — try a different substitution.

4

Integrate in u, back-substitute

Evaluate ∫f(u) du using the basic antiderivative table, then replace u with g(x) to express the answer in x.

Definite Integrals

Change the limits of integration when u-substituting in a definite integral: if u = g(x), the new limits are u = g(a) and u = g(b). Do not back-substitute — evaluate directly in u.

Worked Example

∫6x²(x³+1)⁴ dx
u = x³+1, du = 3x² dx
= 2∫u⁴ du = 2u⁵/5 + C
= 2(x³+1)⁵/5 + C

Integration by Parts

Integration by parts is the product rule in reverse. Use it when the integrand is a product of two different types of functions.

∫u dv = uv − ∫v du

The LIATE Rule — Choosing u

Choose u to be whichever function type appears earliest in the LIATE hierarchy. The remaining factor becomes dv.

L

Logarithmic

ln x, log₂x

I

Inverse Trig

arctan x, arcsin x

A

Algebraic

x², 3x+1, √x

T

Trigonometric

sin x, cos x, tan x

E

Exponential

eˣ, 5ˣ

Worked Examples — LIATE

∫x ln x dx  (L before A → u = ln x)

dv = x dx → v = x²/2

= (x²/2) ln x − ∫(x²/2)(1/x) dx = (x²/2) ln x − x²/4 + C

∫x eˣ dx  (A before E → u = x)

dv = eˣ dx → v = eˣ

= xeˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x − 1) + C

∫arctan x dx  (I before A → u = arctan x)

dv = dx → v = x

= x arctan x − ∫x/(1+x²) dx = x arctan x − (1/2) ln(1+x²) + C

Tabular Method (DI Method)

When you need integration by parts multiple times and u is a polynomial, the tabular method is dramatically faster. Create a D column (differentiate u until 0) and an I column (integrate dv repeatedly). Multiply diagonally with alternating signs.

Example: ∫x³ eˣ dx

SignD (differentiate)I (integrate)
+
3x²
+6x
6
+0

Answer: x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C = eˣ(x³ − 3x² + 6x − 6) + C

Trigonometric Integrals

These strategies handle integrals of the form ∫sinᵐx cosⁿx dx and ∫tanᵐx secⁿx dx by splitting off one factor and applying a Pythagorean identity.

Powers of sin x and cos x

m odd (sin power is odd)

Save one sin x, convert the rest via sin²x = 1 − cos²x, substitute u = cos x.

∫sin³x cos²x dx = ∫(1−cos²x)cos²x sin x dx; u = cos x → −∫(1−u²)u² du

n odd (cos power is odd)

Save one cos x, convert the rest via cos²x = 1 − sin²x, substitute u = sin x.

∫sin²x cos³x dx = ∫sin²x(1−sin²x) cos x dx; u = sin x → ∫u²(1−u²) du

Both m and n even

Use half-angle identities: sin²x = (1 − cos 2x)/2, cos²x = (1 + cos 2x)/2. Reduces to integrals of cos 2x.

∫cos²x dx = ∫(1+cos 2x)/2 dx = x/2 + sin 2x/4 + C

Powers of tan x and sec x

Key identity: tan²x = sec²x − 1. Use it to convert between powers of tan and sec.

n even (sec power is even)

Save sec²x, convert the rest via sec²x = 1 + tan²x, substitute u = tan x.

∫tan³x sec²x dx; u = tan x, du = sec²x dx → ∫u³ du = u⁴/4 + C = tan⁴x/4 + C

m odd (tan power is odd)

Save sec x tan x, convert remaining tan²x using tan²x = sec²x − 1, substitute u = sec x.

∫tan³x sec x dx = ∫(sec²x−1) sec x tan x dx; u = sec x → ∫(u²−1) du

Special cases

∫tan x dx = ln|sec x| + C, ∫sec x dx = ln|sec x + tan x| + C (memorize these).

∫tan²x dx = ∫(sec²x − 1) dx = tan x − x + C

Trigonometric Substitution

Trig substitution eliminates square roots of the form √(a²−x²), √(a²+x²), or √(x²−a²) by converting them to a trig expression without a radical. After integrating, use a right triangle to convert back to x.

√(a² − x²)

Sub: x = a sin θ

dx = a cos θ dθ

√(a² − x²) = a cos θ

Range: −π/2 ≤ θ ≤ π/2

Opposite = x, Hyp = a, Adj = √(a²−x²)

√(a² + x²)

Sub: x = a tan θ

dx = a sec²θ dθ

√(a² + x²) = a sec θ

Range: −π/2 < θ < π/2

Opposite = x, Adj = a, Hyp = √(a²+x²)

√(x² − a²)

Sub: x = a sec θ

dx = a sec θ tan θ dθ

√(x² − a²) = a tan θ

Range: 0 ≤ θ < π/2 or π ≤ θ < 3π/2

Hyp = x, Adj = a, Opposite = √(x²−a²)

Worked Example

∫ 1 / √(9 − x²) dx

a = 3, form √(a²−x²) → x = 3 sin θ, dx = 3 cos θ dθ

√(9 − 9 sin²θ) = 3 cos θ

∫ (3 cos θ dθ) / (3 cos θ) = ∫ dθ = θ + C

Back-substitute: θ = arcsin(x/3)

Answer: arcsin(x/3) + C

Partial Fraction Decomposition

Partial fractions decomposes a rational function into simpler fractions whose antiderivatives are known (logarithms and arctangents). Use it whenever the integrand is a ratio of polynomials.

1

Check degree — divide if necessary

If the degree of the numerator ≥ degree of the denominator, perform polynomial long division first. Continue with the remainder only.

2

Factor the denominator completely

Factor Q(x) into linear factors (ax + b) and irreducible quadratic factors (ax² + bx + c with b² − 4ac < 0). Repeated factors require multiple terms.

3

Write the partial fraction form

Each distinct linear factor (ax+b) gets A/(ax+b). Each repeated factor (ax+b)ⁿ gets A₁/(ax+b) + A₂/(ax+b)² + … + Aₙ/(ax+b)ⁿ. Each irreducible quadratic gets (Ax+B)/(ax²+bx+c).

4

Solve for the constants

Clear the denominators and either (a) substitute convenient values of x to isolate constants, or (b) expand and match coefficients of like powers of x.

5

Integrate each piece

∫A/(ax+b) dx = (A/a) ln|ax+b| + C. Irreducible quadratics with (Ax+B) in the numerator split into a logarithm part and an arctangent part via completing the square.

Example: ∫ (2x + 1) / (x² − x − 2) dx

Factor: x² − x − 2 = (x−2)(x+1)

(2x+1)/((x−2)(x+1)) = A/(x−2) + B/(x+1)

Multiply through: 2x+1 = A(x+1) + B(x−2)

x = 2: 5 = 3A → A = 5/3

x = −1: −1 = −3B → B = 1/3

∫ [5/(3(x−2)) + 1/(3(x+1))] dx

= (5/3) ln|x−2| + (1/3) ln|x+1| + C

Quick Reference: Decomposition Forms

Factor typePartial fraction form
Distinct linear: (ax+b)A/(ax+b)
Repeated linear: (ax+b)²A/(ax+b) + B/(ax+b)²
Repeated linear: (ax+b)ⁿA₁/(ax+b) + … + Aₙ/(ax+b)ⁿ
Irreducible quadratic: (ax²+bx+c)(Ax+B)/(ax²+bx+c)
Repeated quadratic: (ax²+bx+c)²(Ax+B)/(ax²+bx+c) + (Cx+D)/(ax²+bx+c)²

Improper Integrals

Improper integrals involve either infinite limits of integration (Type I) or integrands with vertical asymptotes within the interval (Type II). Both are evaluated using limits.

Type I — Infinite Limits

One or both limits of integration are ±∞. Replace the infinite limit with a variable and take a limit.

Definitions:

∫ₐ^∞ f(x) dx = lim_(b→∞) ∫ₐ^b f(x) dx

∫₋∞^b f(x) dx = lim_(a→−∞) ∫ₐ^b f(x) dx

∫₋∞^∞ f(x) dx = ∫₋∞^c f dx + ∫c^∞ f dx

Type II — Vertical Asymptotes

The integrand blows up at some point c in [a, b]. Split the integral at c and take one-sided limits.

Definitions:

If discontinuity at b: lim_(t→b⁻) ∫ₐ^t f(x) dx

If discontinuity at a: lim_(t→a⁺) ∫t^b f(x) dx

If discontinuity at interior c: split at c, evaluate both sides

Convergence & Divergence

An improper integral converges if the limit exists and is a finite number. It diverges if the limit is ±∞ or does not exist.

p-Test (∫₁^∞ 1/xᵖ dx)

Converges if p > 1, diverges if p ≤ 1.

p = 2: ∫₁^∞ 1/x² dx = 1 (converges)

p = 1: ∫₁^∞ 1/x dx = ∞ (diverges)

p = 1/2: ∫₁^∞ 1/√x dx = ∞ (diverges)

p-Test (∫₀^1 1/xᵖ dx)

Converges if p < 1, diverges if p ≥ 1.

p = 1/2: ∫₀^1 1/√x dx = 2 (converges)

p = 1: ∫₀^1 1/x dx = ∞ (diverges)

p = 2: ∫₀^1 1/x² dx = ∞ (diverges)

Worked Examples

Type I: ∫₁^∞ e^(−x) dx

= lim_(b→∞) [−e^(−x)]₁^b = lim_(b→∞) (−e^(−b) + e^(−1))

= 0 + 1/e = 1/e   Converges.

Type II: ∫₀^1 1/√x dx

= lim_(t→0⁺) ∫t^1 x^(−1/2) dx = lim_(t→0⁺) [2√x]t^1

= lim_(t→0⁺) (2 − 2√t) = 2 − 0

= 2   Converges.

Type I (diverges): ∫₁^∞ 1/x dx

= lim_(b→∞) [ln x]₁^b = lim_(b→∞) ln b = ∞

Diverges.

Choosing the Right Technique

Work through these questions in order. The first matching criterion determines your approach.

1.

1. Does it match a basic antiderivative?

Yes → use the table directly. Done.

Check for xⁿ, eˣ, trig functions, 1/x, and inverse trig forms.

2.

2. Can you simplify algebraically?

Yes → expand, factor, split the fraction, or apply a trig identity first.

Example: ∫sin 2x dx → use sin 2x = 2 sin x cos x or u = 2x.

3.

3. Is there a composite function with its derivative nearby?

Yes → use u-substitution.

The derivative of the inner function must appear as a factor (up to a constant).

4.

4. Is the integrand a product of two different function types?

Yes → use integration by parts. Apply LIATE to choose u.

If you need IBP more than once and u is a polynomial, switch to the tabular method.

5.

5. Does it involve sinᵐ cosⁿ or tanᵐ secⁿ?

Yes → use trig integral strategies (odd power split or half-angle identities).

Check parity of each power to pick the right sub: u = cos x or u = sin x.

6.

6. Does the integrand contain √(a²−x²), √(a²+x²), or √(x²−a²)?

Yes → use trigonometric substitution (sin, tan, or sec substitution).

Complete the square first if needed to get the integral into standard form.

7.

7. Is it a rational function (polynomial / polynomial)?

Yes → use partial fraction decomposition.

Long-divide first if numerator degree ≥ denominator degree.

8.

8. Is there an infinite limit or a vertical asymptote inside the interval?

Yes → it is an improper integral. Rewrite as a limit and evaluate.

If both endpoints are problematic, split the integral at a convenient interior point.

Comprehensive Worked Examples

Example 1 — U-Substitution: ∫ x / (x² + 4) dx

Let u = x² + 4, then du = 2x dx, so x dx = du/2

∫ x/(x²+4) dx = (1/2) ∫ 1/u du = (1/2) ln|u| + C

= (1/2) ln(x² + 4) + C

Note: x² + 4 > 0 always, so no absolute value needed.

Example 2 — Integration by Parts: ∫ x² sin x dx

x² is Algebraic, sin x is Trig → LIATE: u = x². Use tabular method.

D: x², 2x, 2, 0    I: sin x, −cos x, −sin x, cos x

Signs: +, −, +, −

= x²(−cos x) − 2x(−sin x) + 2(cos x) + C

= −x² cos x + 2x sin x + 2 cos x + C

Example 3 — Trig Integral: ∫ sin⁵x dx

m = 5 is odd → save one sin x, convert the rest using sin²x = 1 − cos²x

= ∫ (1 − cos²x)² sin x dx

Let u = cos x, du = −sin x dx

= −∫ (1 − u²)² du = −∫ (1 − 2u² + u⁴) du

= −u + (2/3)u³ − (1/5)u⁵ + C

= −cos x + (2/3) cos³x − (1/5) cos⁵x + C

Example 4 — Trig Substitution: ∫ √(4 − x²) dx

a = 2, form √(a²−x²) → x = 2 sin θ, dx = 2 cos θ dθ

√(4 − 4 sin²θ) = 2 cos θ

∫ 2 cos θ · 2 cos θ dθ = 4 ∫ cos²θ dθ

= 4 ∫ (1 + cos 2θ)/2 dθ = 2θ + sin 2θ + C

= 2θ + 2 sin θ cos θ + C

Back-sub: θ = arcsin(x/2), sin θ = x/2, cos θ = √(4−x²)/2

= 2 arcsin(x/2) + (x/2)√(4−x²) + C

Example 5 — Partial Fractions: ∫ 1 / (x² − 1) dx

Factor: x² − 1 = (x−1)(x+1)

1/((x−1)(x+1)) = A/(x−1) + B/(x+1)

1 = A(x+1) + B(x−1)

x = 1: 1 = 2A → A = 1/2   x = −1: 1 = −2B → B = −1/2

∫ [(1/2)/(x−1) − (1/2)/(x+1)] dx

= (1/2) ln|x−1| − (1/2) ln|x+1| + C = (1/2) ln|(x−1)/(x+1)| + C

Frequently Asked Questions

How do you know when to use u-substitution?

Use u-substitution when you can identify a composite function inside the integral and its derivative (or a constant multiple of it) also appears in the integrand. Look for a function and its derivative together: ∫f(g(x))g'(x) dx. Set u = g(x) (the inner function), compute du = g'(x) dx, and substitute to simplify. Example: ∫2x·cos(x²) dx — let u = x², du = 2x dx, transforms to ∫cos(u) du = sin(u) + C = sin(x²) + C.

What is the LIATE rule for integration by parts?

LIATE is a mnemonic for choosing u in integration by parts (∫u dv = uv − ∫v du). Choose u to be the type of function that appears earliest in the acronym: L = Logarithmic (ln x, log x), I = Inverse trig (arctan x, arcsin x), A = Algebraic (polynomials like x², 3x + 1), T = Trigonometric (sin x, cos x), E = Exponential (eˣ, 2ˣ). Example: ∫x·eˣ dx — x is Algebraic and eˣ is Exponential; A comes before E, so u = x and dv = eˣ dx. Result: xeˣ − eˣ + C.

When do you use the tabular method for integration by parts?

Use the tabular method (also called the DI method or repeated integration by parts) when you must apply integration by parts multiple times and u is a polynomial. Create two columns: D (differentiate u repeatedly until reaching 0) and I (integrate dv repeatedly). Alternate the signs +, −, +, −, … and multiply diagonally. Example: ∫x³eˣ dx — D column: x³, 3x², 6x, 6, 0; I column: eˣ, eˣ, eˣ, eˣ, eˣ. Answer: x³eˣ − 3x²eˣ + 6xeˣ − 6eˣ + C = eˣ(x³ − 3x² + 6x − 6) + C.

How do you integrate powers of sine and cosine?

For ∫sinᵐx cosⁿx dx: If m is odd, save one sin x, convert the rest using sin²x = 1 − cos²x, then substitute u = cos x. If n is odd, save one cos x, convert the rest using cos²x = 1 − sin²x, then substitute u = sin x. If both m and n are even, use half-angle identities: sin²x = (1 − cos 2x)/2 and cos²x = (1 + cos 2x)/2. Example: ∫sin³x dx = ∫sin²x · sin x dx = ∫(1 − cos²x) sin x dx; let u = cos x, du = −sin x dx → −∫(1 − u²) du = −cos x + cos³x/3 + C.

What are the three trigonometric substitutions and when do you use each?

The three trig substitutions match the three Pythagorean identities. (1) For √(a² − x²): substitute x = a sin θ, dx = a cos θ dθ. The expression becomes a cos θ. Use when the integrand contains a² − x². (2) For √(a² + x²): substitute x = a tan θ, dx = a sec² θ dθ. The expression becomes a sec θ. Use when the integrand contains a² + x². (3) For √(x² − a²): substitute x = a sec θ, dx = a sec θ tan θ dθ. The expression becomes a tan θ. Use when the integrand contains x² − a². Always draw a right triangle to convert back to x at the end.

How does partial fraction decomposition work?

Partial fraction decomposition rewrites a rational function P(x)/Q(x) as a sum of simpler fractions that are easy to integrate. Step 1: If degree of P ≥ degree of Q, perform polynomial long division first. Step 2: Factor the denominator Q(x) completely. Step 3: Write a partial fraction for each factor — A/(ax+b) for linear factors, (Ax+B)/(ax²+bx+c) for irreducible quadratics, and repeated factors require extra terms like A/(ax+b) + B/(ax+b)². Step 4: Clear denominators and solve for the constants by substituting convenient values of x or matching coefficients. Step 5: Integrate each partial fraction — linear factors give logarithms, irreducible quadratics give arctangents.

What is an improper integral and how do you determine if it converges?

An improper integral has either an infinite limit of integration (Type I) or an integrand with a vertical asymptote in the interval (Type II). To evaluate, replace the problematic part with a limit. Type I example: ∫₁^∞ 1/x² dx = lim_{b→∞} ∫₁^b x⁻² dx = lim_{b→∞} [−1/x]₁^b = lim_{b→∞} (−1/b + 1) = 1. This converges. Type II example: ∫₀^1 1/√x dx = lim_{a→0⁺} ∫ₐ^1 x^{−1/2} dx = lim_{a→0⁺} [2√x]ₐ^1 = 2. This converges. If the limit does not exist or is ±∞, the integral diverges. The p-test: ∫₁^∞ 1/xᵖ dx converges if p > 1 and diverges if p ≤ 1.

How do you choose the right integration technique for a given integral?

Follow this decision process: (1) Is it a basic form? Check the antiderivative table first — many integrals are done immediately. (2) Can you simplify? Expand products, split fractions, use trig identities before trying a technique. (3) Is there a composite function with its derivative? Use u-substitution. (4) Is it a product of two different function types? Use integration by parts — apply LIATE to choose u. (5) Does it involve sin^m · cos^n, tan^m · sec^n, etc.? Use trig integral strategies. (6) Does it contain √(a²−x²), √(a²+x²), or √(x²−a²)? Use trig substitution. (7) Is it a rational function (polynomial over polynomial)? Use partial fractions. (8) Does it have an infinite bound or vertical asymptote? It is an improper integral — convert to a limit.

What is the reduction formula for integration by parts with trig powers?

Reduction formulas reduce the power of a trig function by 2 each time, eventually arriving at a base case. For sine: ∫sinⁿx dx = −(1/n)sinⁿ⁻¹x cos x + ((n−1)/n)∫sinⁿ⁻²x dx. For cosine: ∫cosⁿx dx = (1/n)cosⁿ⁻¹x sin x + ((n−1)/n)∫cosⁿ⁻²x dx. For secant (very useful in Calc 2): ∫secⁿx dx = (1/(n−1))secⁿ⁻²x tan x + ((n−2)/(n−1))∫secⁿ⁻²x dx. Apply the formula repeatedly until reaching ∫sin x dx, ∫cos x dx, ∫sec x dx, or ∫sec²x dx — all of which are standard antiderivatives.

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