Tags: integration by parts, IBP, partial fractions, trig integrals, trigonometric integration, sec integral, arctan integral, u-substitution, Calculus II, Texas A&M, MATH 172, MATH 2414
This covers the main integration methods you need for Calc II: integration by parts, trigonometric integrals (odd/even powers of sin and cos, secant), partial fraction decomposition, and recognising standard integral forms like arctan. These come up constantly on exams, both as standalone questions and as steps inside larger problems (volumes, arc length, work).
Integration by parts (IBP)
A technique for integrating products of functions, based on the product rule in reverse. Formula: ∫u dv = uv − ∫v du.
Partial fraction decomposition
A method for integrating rational functions by breaking them into simpler fractions whose denominators are linear or irreducible quadratic factors.
Trigonometric integral
An integral involving powers of sin, cos, sec, csc, tan, or cot. Strategy depends on whether the powers are odd or even.
u-substitution
Reversing the chain rule to simplify an integral by substituting u = g(x).
The formula is ∫u dv = uv − ∫v du.
Choose u and dv so that ∫v du is simpler than the original integral
Common mnemonic for choosing u: LIATE (Logarithmic, Inverse trig, Algebraic, Trig, Exponential)
When ln(x) appears in the integrand, it nearly always becomes u
Worked example: ∫x ln(x) dx
Let u = ln(x), dv = x dx
Then du = (1/x) dx, v = x²/2
∫x ln(x) dx = (x²/2) ln(x) − ∫(x²/2)(1/x) dx
= (x²/2) ln(x) − (1/2)∫x dx
= (x²/2) ln(x) − x²/4 + C
Odd power of cosine (or sine): peel one off, convert the rest.
When cos has an odd power, save one cos(x) for the du and convert remaining cos² to (1 − sin²).
Worked example: ∫cos³(x) sin²(x) dx
Rewrite as ∫cos²(x) sin²(x) cos(x) dx
Replace cos²(x) = 1 − sin²(x)
= ∫(1 − sin²(x)) sin²(x) cos(x) dx
Let u = sin(x), du = cos(x) dx
= ∫(1 − u²)u² du = ∫(u² − u⁴) du
= u³/3 − u⁵/5 + C
= sin³(x)/3 − sin⁵(x)/5 + C
The same logic works in reverse: if sin has an odd power, peel one sin(x) off and convert sin² to (1 − cos²).
The integral of sec(x)
∫sec(x) dx = ln|sec(x) + tan(x)| + C
This is a standard result worth memorising. It comes from multiplying numerator and denominator by (sec(x) + tan(x)).
Used when integrating a rational function P(x)/Q(x) where deg(P) < deg(Q).
Factor the denominator completely
Write one fraction per factor: A/(linear factor), (Ax + B)/(irreducible quadratic)
Solve for coefficients by plugging in convenient x values or equating coefficients
Integrate each fraction separately
Worked example: ∫1/(x² + 3x + 2) dx
Factor: x² + 3x + 2 = (x + 1)(x + 2)
Decompose: 1/((x+1)(x+2)) = A/(x+1) + B/(x+2)
Multiply through: 1 = A(x + 2) + B(x + 1)
Set x = −1: A = 1
Set x = −2: B = −1
∫(1/(x+1) − 1/(x+2)) dx = ln|x + 1| − ln|x + 2| + C
These are forms you should recognise on sight:
∫1/(x² + 1) dx = arctan(x) + C
∫1/√(1 − x²) dx = arcsin(x) + C
∫sec(x) dx = ln|sec(x) + tan(x)| + C
More generally, ∫1/(x² + a²) dx = (1/a) arctan(x/a) + C.
Integral | Result |
|---|---|
∫u dv | uv − ∫v du |
∫sec(x) dx | ln|sec(x) + tan(x)| + C |
∫1/(x² + 1) dx | arctan(x) + C |
∫1/(x² + a²) dx | (1/a) arctan(x/a) + C |
Trig identity for odd-power integrals: cos²(x) = 1 − sin²(x), and sin²(x) = 1 − cos²(x).
⚠️ In IBP, forgetting to include the constant of integration or making sign errors when evaluating ∫v du is a common way to lose marks.
⚠️ For trig integrals, the strategy depends on whether the power is odd or even. Odd power: peel and substitute. Even power: use half-angle identities. Mixing these up leads to dead ends.
⚠️ When doing partial fractions, always check that the degree of the numerator is strictly less than the degree of the denominator. If it is not, do polynomial long division first.
⚠️ Do not confuse ∫1/(x² + 1) dx = arctan(x) with ∫1/√(1 − x²) dx = arcsin(x). These two forms are frequently tested side by side.
Q: Evaluate ∫x ln(x) dx.
A: (x²/2) ln(x) − x²/4 + C. Use IBP with u = ln(x), dv = x dx.
Q: Evaluate ∫cos³(x) sin²(x) dx.
A: sin³(x)/3 − sin⁵(x)/5 + C. Peel off one cos(x), convert cos²(x) = 1 − sin²(x), then substitute u = sin(x).
Q: Evaluate ∫1/(x² + 3x + 2) dx using partial fractions.
A: ln|x + 1| − ln|x + 2| + C. Factor denominator as (x+1)(x+2), decompose, and integrate.
Q: What is ∫sec(x) dx?
A: ln|sec(x) + tan(x)| + C.
Q: What is ∫1/(x² + 1) dx?
A: arctan(x) + C.
integration by parts, IBP, LIATE, tabular integration, partial fractions, partial fraction decomposition, trigonometric integrals, trig integrals, odd power cosine, odd power sine, secant integral, arctan integral, inverse trig integrals, u-substitution, Calc II integration methods, MATH 2414, MATH 172