Difficulty: Intermediate | Prerequisites: Differential calculus (derivatives, chain rule), basic antiderivatives, familiarity with exponential and trigonometric functions.
Integration techniques are the toolkit you reach for once you know what an integral is but the integrand does not yield to a simple antiderivative. This material sits at the core of any integral calculus course: every later topic (areas, volumes, differential equations, series) assumes you can evaluate integrals fluently. You should already be comfortable differentiating composite, exponential and trigonometric functions, because several of these methods (substitution, the Fundamental Theorem with chain rule) are differentiation run in reverse.
The Fundamental Theorem of Calculus links definite integrals to antiderivatives. When the upper limit is itself a function, the chain rule comes in. u-substitution, integration by parts and product-to-sum trig identities each handle a different class of integrand, and recognising which tool fits is half the work.
Fundamental Theorem of Calculus (FTC)
If f is continuous on [a, b] and F is any antiderivative of f, then the definite integral from a to b of f(x) dx equals F(b) minus F(a). In simple terms, this means you can evaluate a definite integral by finding an antiderivative and plugging in the limits.
FTC, Part 1 (derivative of an integral with variable upper limit)
If g(x) is the integral from a to x of f(t) dt, then g'(x) = f(x). Think of it as: differentiating undoes the integral, giving back the integrand evaluated at the upper limit.
Chain Rule Extension of FTC
When the upper limit is a function u(x) rather than x itself, the derivative picks up the factor u'(x). Formally: d/dx of the integral from a to u(x) of f(t) dt equals f(u(x)) times u'(x).
u-Substitution
A technique that reverses the chain rule. You set u equal to an inner function, replace dx with du, and transform the integral into a simpler one. Think of it as swapping variables so the integral looks like something you already know.
Integration by Parts
A technique that reverses the product rule. The formula is: integral of u dv equals uv minus the integral of v du. You choose u and dv so that the new integral (v du) is easier than the original.
Product-to-Sum Identities (Trigonometric)
Identities that convert a product of sines and cosines into a sum or difference: sin A sin B = (1/2)(cos(A minus B) minus cos(A plus B)), and similar forms. In simple terms, these turn a tricky trig product into two straightforward cosine or sine terms you can integrate directly.
When F(x) is defined as an integral whose upper limit is a function of x, differentiate in two steps:
First, let f(x) be the integral from a constant to x of the integrand. By FTC Part 1, f'(x) equals the integrand evaluated at x.
Then apply the chain rule: F'(x) = f'(u(x)) times u'(x).
Example from exam: F(x) = integral from 0 to sqrt(x) of (12 e^(-t^2))/(4 + 3t^2) dt.
Set f(x) = integral from 0 to x of the same integrand, so f'(x) = (12 e^(-x^2))/(4 + 3x^2).
Since F(x) = f(sqrt(x)), the chain rule gives F'(x) = f'(sqrt(x)) times 1/(2 sqrt(x)) = (6 e^(-x)) / (sqrt(x)(4 + 3x)).
Evaluating at x = 4: F'(4) = (3/16) e^(-4).
When the integrand is a product that can be expanded into simpler terms, distribute first, then integrate term by term.
Example: integral from 1 to 4 of sqrt(x)(3 + 4/x) dx.
Expand: 3 sqrt(x) + 4/sqrt(x).
Integrate: 2x^(3/2) + 8x^(1/2), then evaluate between 1 and 4 to get 22.
Identify an inner function u = g(x) whose derivative g'(x) appears (or nearly appears) as a factor in the integrand.
Replace all x-expressions with u and dx with du/g'(x).
For definite integrals, convert the limits as well.
Example: integral from 0 to 1 of (4 arctan(x))/(1 + x^2) dx.
Set u = arctan(x), so du = dx/(1 + x^2).
Limits: x = 0 gives u = 0, x = 1 gives u = pi/4.
The integral becomes integral from 0 to pi/4 of 4u du = 2u^2 evaluated from 0 to pi/4 = pi^2/8.
Example: area bounded by f(x) = (6 ln x)/x, y = 0, and x = 2.
The lower limit is where f(x) = 0, i.e. x = 1.
Set u = ln x, du = dx/x. Limits: x = 1 gives u = 0, x = 2 gives u = ln 2.
Integral becomes integral from 0 to ln 2 of 6u du = 3(ln 2)^2.
Use when the integrand is a product of two functions where one simplifies on differentiation and the other is easy to integrate (or vice versa).
Formula: integral of u dv = uv minus integral of v du.
Choosing u and dv well is the key skill. Look for derivative relationships that simplify the new integral.
Example: integral from 0 to 1 of (4x e^x)/(x + 1)^2 dx.
Notice d/dx(1/(x+1)) = -1/(x+1)^2 and d/dx(xe^x) = e^x(x+1).
Set u-part = 4xe^x, dv-part = dx/(x+1)^2.
After applying parts: [-4xe^x/(x+1)] from 0 to 1 + 4 times integral from 0 to 1 of e^x dx = -2e + 4e - 4 = 2e - 4.
For integrals of the form sin(Ax) sin(Bx), cos(Ax) cos(Bx) or sin(Ax) cos(Bx), convert to sums first.
sin A sin B = (1/2)(cos(A - B) - cos(A + B)).
Example: integral of 4 sin(5x) sin(2x) dx.
Apply identity: 4 times (1/2)(cos 3x - cos 7x) = 2(cos 3x - cos 7x).
Integrate: (2/3) sin 3x - (2/7) sin 7x + C.
\frac{d}{dx} \int_a^{u(x)} f(t)\,dt = f(u(x)) \cdot u'(x)\int u\,dv = uv - \int v\,du\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]u-substitution and integration by parts appear constantly in physics and engineering. Any time you compute work done by a variable force, the charge stored in a capacitor, or the probability under a continuous distribution, you are evaluating an integral that rarely has a textbook-simple antiderivative. Product-to-sum identities turn up in signal processing, where multiplying two sine waves (modulation) produces sum and difference frequencies.
Forgetting the chain rule factor when the upper limit is not simply x. If the upper limit is sqrt(x), you must multiply by 1/(2 sqrt(x)). Many students write f'(sqrt(x)) alone and lose marks.
Confusing the sign in the product-to-sum identity for sin A sin B. The result is cos(A - B) MINUS cos(A + B), not a sum. Mixing up the sign flips the entire answer.
Applying integration by parts with a poor choice of u and dv, producing a harder integral rather than a simpler one. If the new integral is more complicated, try swapping your choices.
Forgetting to change the limits of integration when using u-substitution on a definite integral. If you switch to the variable u, the limits must be expressed in u as well, or you must convert back to x before evaluating.
FTC + chain rule problems appear on nearly every integral calculus exam. Expect the upper limit to be sqrt(x), x^2, or similar.
u-substitution is the single most-tested technique. If you see 1/(1 + x^2) paired with arctan, or (1/x) paired with ln x, substitution is almost certainly the intended method.
Integration by parts questions often require spotting a derivative relationship between two factors. The exam may pair x e^x with 1/(x+1)^2, as here.
Trig product integrals test whether you know the product-to-sum identities. These are quick marks if you have the identities memorised, and nearly impossible otherwise.
True or False: If F(x) = integral from 0 to x^2 of sin(t) dt, then F'(x) = sin(x^2). _____ (False: F'(x) = sin(x^2) times 2x, by the chain rule.)
Fill in the blank: sin A sin B = (1/2)(cos() minus cos()). (A minus B; A plus B)
True or False: When using u-substitution on a definite integral, you may keep the original x-limits if you convert back to x before evaluating. _____ (True.)
Fill in the blank: The integration by parts formula is: integral of u dv = ____ minus ____. (uv; integral of v du)
True or False: If d/dx(xe^x) = e^x(x + 1), then a good pairing for integrating (xe^x)/(x+1)^2 by parts is to let the xe^x factor be the part you integrate. _____ (False: xe^x pairs with 1/(x+1)^2 as u times dv, where dv = dx/(x+1)^2.)
Q: If F(x) = integral from 0 to sqrt(x) of (12 e^(-t^2))/(4 + 3t^2) dt, what is F'(4)?
A: Set f(x) = integral from 0 to x of the integrand. By FTC, f'(x) = (12 e^(-x^2))/(4 + 3x^2). Since F(x) = f(sqrt(x)), the chain rule gives F'(x) = f'(sqrt(x))/(2 sqrt(x)) = (6 e^(-x))/(sqrt(x)(4 + 3x)). At x = 4: F'(4) = (3/16) e^(-4).
Q: Evaluate the integral from 1 to 4 of sqrt(x)(3 + 4/x) dx.
A: Expand to 3 sqrt(x) + 4/sqrt(x). Antiderivatives are 2x^(3/2) + 8x^(1/2). Evaluate: (2 times 8 + 8 times 2) minus (2 + 8) = 32 minus 10 = 22.
Q: Evaluate the integral from 0 to 1 of (4 arctan x)/(1 + x^2) dx.
A: Substitute u = arctan x, du = dx/(1 + x^2). Limits become 0 to pi/4. The integral is integral of 4u du from 0 to pi/4 = 2u^2 evaluated at pi/4 = pi^2/8.
Q: Find the integral of 4 sin(5x) sin(2x) dx.
A: Use sin A sin B = (1/2)(cos(A-B) - cos(A+B)). The integral becomes 2 times integral of (cos 3x - cos 7x) dx = (2/3) sin 3x - (2/7) sin 7x + C.
Q: Evaluate the integral from 0 to 1 of (4x e^x)/(x+1)^2 dx.
A: Use integration by parts with u = 4xe^x and dv = dx/(x+1)^2 (since d/dx(1/(x+1)) = -1/(x+1)^2 and d/dx(xe^x) = e^x(x+1)). Result: [-4xe^x/(x+1)] from 0 to 1 + 4 integral of e^x dx = -2e + 4(e - 1) = 2e - 4.
This connects to improper integrals, where you will apply these same techniques but with infinite limits or discontinuous integrands. It also connects to sequences and series, because the Integral Test for convergence requires evaluating an improper integral of the series' underlying function. Volumes of revolution and area computations are direct applications of definite integrals evaluated with these methods.
Fundamental Theorem of Calculus, FTC, chain rule for integrals, u-substitution, change of variable, integration by parts, IBP, product-to-sum formula, trig product integrals, definite integral evaluation, antiderivative, MATH 101, integral calculus, UT Austin