Difficulty: Intermediate to Advanced | Prerequisites: Basic integration rules, chain rule, trig identities (Pythagorean identity)
Calc II is, in large part, a course about learning to evaluate integrals that basic antidifferentiation rules cannot handle. This problem set covers four core techniques: u-substitution for composed functions, integration by parts (including the recursive/reduction formula pattern), trigonometric integrals with odd or even powers, and straightforward algebraic substitution. Mastering these four covers the majority of integration problems you will face on exams.
U-substitution rewrites a composed integral in terms of a simpler variable. Integration by parts trades one integral for another (and sometimes requires repeated application via a reduction formula). Trigonometric integrals with an odd power use the Pythagorean identity to peel off one factor and then substitute. Simple algebraic substitution handles integrals where the denominator or exponent suggests an obvious new variable.
U-substitution
Replacing part of the integrand with a new variable u so that the integral simplifies. You also rewrite dx in terms of du using du = u'(x) dx. Think of it as the chain rule running backwards.
Integration by parts
A technique based on the product rule: integral of u dv = uv minus integral of v du. You choose which part of the integrand is u and which is dv. In simple terms, you trade the original integral for a (hopefully simpler) one.
Reduction formula / recursion
A formula that expresses the integral of u^n e^u du (or similar) in terms of the integral of u^(n-1) e^u du. You apply it repeatedly until you reach a base case. Think of it as peeling off one power at a time.
Trigonometric integral
An integral involving powers of sin and cos (or other trig functions). The strategy depends on whether the powers are odd or even. For odd powers, strip off one factor and use the Pythagorean identity to convert the rest.
Pythagorean identity
sin^2(theta) + cos^2(theta) = 1. Rearranged: sin^2(theta) = 1 - cos^2(theta) or cos^2(theta) = 1 - sin^2(theta). This is the workhorse identity for converting between sin and cos in trig integrals.
When the integrand has the form x^n sin(x^k) cos^m(x^k), look for a substitution that simplifies the trig composition.
Worked Example (PS1, 3a): integral of x^3 sin(x^4) cos^7(x^4) dx
Spot the inner function: x^4 appears inside both sin and cos
Let u = cos(x^4), then du = -4x^3 sin(x^4) dx
Rearrange: x^3 sin(x^4) dx = (-1/4) du
The integral becomes: (-1/4) integral of u^7 du
Integrate: (-1/4)(u^8 / 8) + C = (-1/32) u^8 + C
Substitute back: (-1/32) cos^8(x^4) + C
The key insight is recognising that x^3 sin(x^4) dx is (up to a constant) the differential of cos(x^4).
When the integrand is a polynomial times an exponential (or similar), integration by parts must be applied repeatedly. The reduction formula for integral of u^n e^u du is:
integral of u^n e^u du = u^n e^u - n times integral of u^(n-1) e^u du
Worked Example (PS1, 3b): integral from 0 to 1 of x^(3/2) e^(sqrt(x)) dx
First, simplify with substitution: let u = sqrt(x), so x = u^2, dx = 2u du
When x = 0, u = 0; when x = 1, u = 1
x^(3/2) = (u^2)^(3/2) = u^3, and e^(sqrt(x)) = e^u
The integral becomes: 2 times integral from 0 to 1 of u^4 e^u du
Apply the reduction formula four times, starting from n = 4:
integral of u^4 e^u = u^4 e^u - 4 times integral of u^3 e^u
integral of u^3 e^u = u^3 e^u - 3 times integral of u^2 e^u
integral of u^2 e^u = u^2 e^u - 2 times integral of u e^u
integral of u e^u = u e^u - e^u
Combine: integral of u^4 e^u = e^u(u^4 - 4u^3 + 12u^2 - 24u + 24)
Evaluate from 0 to 1: [e(1 - 4 + 12 - 24 + 24)] - [1(0 - 0 + 0 - 0 + 24)] = 9e - 24
Final answer: 2(9e - 24) = 18e - 48
When one of sin or cos has an odd power, peel off one factor of that function, convert the remaining even power using the Pythagorean identity, then substitute.
Worked Example (PS1, 3c): integral of sin^9(theta) cos^4(theta) d(theta)
sin has the odd power (9). Write sin^9 = sin^8 times sin = (sin^2)^4 times sin
Replace sin^2 = 1 - cos^2: the integrand becomes (1 - cos^2)^4 cos^4 sin d(theta)
Let u = cos(theta), du = -sin(theta) d(theta)
The integral becomes: -integral of (1 - u^2)^4 u^4 du
Expand (1 - u^2)^4 = 1 - 4u^2 + 6u^4 - 4u^6 + u^8
Multiply by u^4: u^4 - 4u^6 + 6u^8 - 4u^10 + u^12
Integrate term by term: -(u^5/5 - 4u^7/7 + 6u^9/9 - 4u^11/11 + u^13/13) + C
Substitute back u = cos(theta)
Result: -(1/13)cos^13(theta) + (4/11)cos^11(theta) - (2/3)cos^9(theta) + (4/7)cos^7(theta) - (1/5)cos^5(theta) + C
When the integrand contains an expression and its derivative (or close to it), a direct substitution cleans things up immediately.
Worked Example (PS1, 3d): integral of 2x / (x^2 + 1)^(3/2) dx
Notice that d/dx(x^2 + 1) = 2x, which is the numerator
Let u = x^2 + 1, du = 2x dx
The integral becomes: integral of u^(-3/2) du
Integrate: u^(-1/2) / (-1/2) + C = -2 u^(-1/2) + C
Substitute back: -2 / sqrt(x^2 + 1) + C
The problem set explicitly notes that trig substitution or hyperbolic substitution would receive no credit here, because the simple u-sub is the intended (and fastest) method.
Integration by parts:
\int u\,dv = uv - \int v\,duReduction formula for polynomial times exponential:
\int u^n e^u\,du = u^n e^u - n\int u^{n-1} e^u\,duPythagorean identity (the trig integral workhorse):
\sin^2\theta = 1 - \cos^2\theta \qquad \cos^2\theta = 1 - \sin^2\thetaPower rule for integration:
\int u^n\,du = \frac{u^{n+1}}{n+1} + C \quad (n \neq -1)Students often try to use trig substitution (x = tan(theta)) or hyperbolic substitution on integrals that have a clean algebraic u-sub. Always check for a simple substitution first; it saves time and, as Problem 3d shows, the professor may explicitly penalise the longer route.
When applying integration by parts recursively, students lose track of signs. Each application introduces a minus sign, and errors compound. Write out each step carefully.
In trig integrals, students sometimes try to use double-angle formulas when the power is odd. That is the strategy for even powers. When one power is odd, the Pythagorean identity approach (peel off one factor, convert the rest) is simpler and more direct.
Students forget the constant of integration on indefinite integrals. On definite integrals, it cancels, but on indefinite integrals its absence costs marks.
U-substitution is the single most-tested integration technique. If you master nothing else, master this.
Integration by parts with recursion often appears as a longer, multi-step problem worth significant marks. Practise the reduction formula until it is automatic.
Trig integrals with odd powers appear frequently. The even-power case (using half-angle identities) is also important but was not tested in this problem set.
When the problem says "no credit" for a particular method, pay attention. Professors test whether you can identify the simplest approach, not just any approach that works.
Always simplify your final answer. Leaving cos(x^4) inside a power is fine, but leaving unsimplified fractions or un-substituted u-variables is not.
True or False: When one trig function has an odd power, use half-angle formulas. (False, use Pythagorean identity)
Fill in the blank: The reduction formula for integral of u^n e^u du is u^n e^u minus ____ times integral of ____. (n; u^(n-1) e^u du)
True or False: For integral of 2x / (x^2 + 1)^(3/2) dx, the fastest method is trig substitution. (False, direct u-sub is faster and was required)
Fill in the blank: When using u-sub, you must rewrite dx in terms of ____. (du)
Q: Evaluate integral of x^2 cos(x^3) dx.
A: Let u = x^3, du = 3x^2 dx, so x^2 dx = du/3. The integral becomes (1/3) integral of cos(u) du = (1/3) sin(u) + C = (1/3) sin(x^3) + C.
Q: Evaluate integral of sin^5(x) cos^2(x) dx.
A: sin has the odd power. Write sin^5 = (sin^2)^2 sin = (1 - cos^2)^2 sin. Let u = cos(x), du = -sin(x) dx. The integral becomes -integral of (1 - u^2)^2 u^2 du. Expand, integrate term by term, substitute back.
Q: Why does the reduction formula work for integral of u^n e^u du?
A: It is just integration by parts with the specific choice: let the "u" in IBP be u^n (which simplifies when differentiated) and let dv = e^u du (which stays the same when integrated). Each application reduces the power by 1.
Q: Evaluate integral of 6x^2 / (x^3 + 5)^4 dx.
A: Let u = x^3 + 5, du = 3x^2 dx, so 6x^2 dx = 2 du. The integral becomes 2 integral of u^(-4) du = 2 times u^(-3)/(-3) + C = -2 / (3(x^3 + 5)^3) + C.
These techniques are the toolkit for the rest of Calc II. Integration by parts returns when deriving formulas for Taylor series and when evaluating improper integrals. Trig integrals appear in Fourier analysis and in computing arc lengths and surface areas of revolution. U-substitution is everywhere.
The trig substitution and hyperbolic substitution methods mentioned (but disallowed) in Problem 3d are themselves important techniques covered later in the course for integrals involving sqrt(a^2 - x^2), sqrt(x^2 + a^2), and sqrt(x^2 - a^2).
U-substitution, integration by parts, IBP, reduction formula, recursive integration, tabular integration, trig integrals, trigonometric integrals odd power, Pythagorean identity substitution, sin^n cos^m integrals, algebraic substitution, Calc 2 NYU, antiderivative techniques, integral of polynomial times exponential