Source: Lecture 3 Examples
Tags: basic integration, u-substitution, integration by parts, arctan integral, FTC, chain rule, tangent line, Calculus II, NYU
Difficulty: Intermediate | Prerequisites: Calculus I (differentiation, basic antiderivatives, Fundamental Theorem of Calculus)
This material sits at the start of Calculus II, bridging the gap between differentiation and the more advanced integration techniques that dominate the rest of the course. Before you can tackle trigonometric integrals, partial fractions or improper integrals, you need fluency with three workhorses: the Fundamental Theorem of Calculus applied through the chain rule, u-substitution for reshaping integrands, and integration by parts for products. If you are comfortable differentiating composite functions and know your basic antiderivative table, you are ready for this.
This lecture covers three core skills: using the FTC plus the chain rule to find derivatives of integral-defined functions (and their tangent lines), applying u-substitution to simplify exponential and rational integrands, and using integration by parts to evaluate definite integrals involving inverse trig functions. Every Calculus II exam leans heavily on these techniques.
Fundamental Theorem of Calculus (FTC), Part 1
If F(x) = the integral from a to x of f(t) dt, then F'(x) = f(x). In plain terms, differentiating an integral whose upper limit is x just gives you back the integrand, evaluated at x.
Chain Rule Extension of the FTC
When the upper limit is a function g(x) rather than plain x, you differentiate as: d/dx of the integral from a to g(x) of f(t) dt = f(g(x)) times g'(x). Think of it as the FTC result, multiplied by the derivative of whatever sits in the upper limit.
U-substitution
A technique for simplifying integrals by replacing a complicated expression with a single variable u. You set u equal to some inner function, compute du, and rewrite the entire integral in terms of u. In simple terms, it is the reverse of the chain rule for differentiation.
Integration by Parts
Based on the product rule in reverse: the integral of u dv = uv minus the integral of v du. You choose which part of the integrand to call u (the part that simplifies when differentiated) and which to call dv (the part you can integrate). Think of it as a way to trade a hard integral for an easier one.
Arctan (Inverse Tangent)
The function arctan(x) gives the angle whose tangent is x. Its derivative is 1/(1 + x squared), which is why it appears frequently in integration problems involving 1/(1 + x squared).
The first example demonstrates how to find the equation of a tangent line to a function defined partly by an integral.
The Setup
Given f(x) = cos(x + pi/4) + 3 times the integral from 1 to (2x squared + x + 1) of log(t + 9) dt, find the tangent line at x = 0.
Step 1: Find f(0)
Evaluate each piece at x = 0.
cos(0 + pi/4) = cos(pi/4) = sqrt(2)/2.
The integral from 1 to 1 of log(t + 9) dt = 0, because the limits are equal.
So f(0) = sqrt(2)/2.
Step 2: Find f'(x) using the chain rule extension of the FTC
Differentiate the cosine piece: the derivative of cos(x + pi/4) is negative sin(x + pi/4).
Differentiate the integral piece using the FTC chain rule: multiply the integrand evaluated at the upper limit, log(2x squared + x + 10), by the derivative of the upper limit, (4x + 1), then multiply by 3.
f'(x) = negative sin(x + pi/4) + (12x + 3) times log(2x squared + x + 10).
Step 3: Evaluate f'(0)
f'(0) = negative sin(pi/4) + 3 times log(10).
f'(0) = negative sqrt(2)/2 + 3 log(10).
Step 4: Write the tangent line
Using point-slope form: y minus sqrt(2)/2 = (negative sqrt(2)/2 + 3 log(10)) times (x minus 0).
Equivalently: y = (negative sqrt(2)/2 + 3 log(10)) times x + sqrt(2)/2.
Two related integrals are solved here, both involving e to the x in the denominator. The key insight is recognising what to set as u.
Problem A: integral of e^x / (1 + e^(2x)) dx
Set u = e^x, so du = e^x dx.
The denominator rewrites: 1 + e^(2x) = 1 + (e^x) squared = 1 + u squared.
The integral becomes: integral of du / (1 + u squared) = arctan(u) + C = arctan(e^x) + C.
This works because the standard result integral of 1/(1 + u squared) du = arctan(u) is one you should have memorised.
Problem B: integral of e^x / (1 + e^x) dx
Set u = e^x, so du = e^x dx.
The integral becomes: integral of du / (1 + u) = ln|1 + u| + C = ln|1 + e^x| + C.
This is a straightforward ln-form antiderivative.
Takeaway
The same substitution, u = e^x, leads to two very different standard forms depending on the power in the denominator. Recognising whether the denominator is 1 + u (giving ln) or 1 + u squared (giving arctan) is the whole game.
This example evaluates the definite integral from 0 to 1 of arctan(x) dx using integration by parts.
Choosing u and dv
Set u = arctan(x), so du = 1/(1 + x squared) dx.
Set dv = dx, so v = x.
Integration by parts gives: x times arctan(x) evaluated from 0 to 1, minus the integral from 0 to 1 of x/(1 + x squared) dx.
Evaluating the boundary term
At x = 1: 1 times arctan(1) = pi/4.
At x = 0: 0 times arctan(0) = 0.
Boundary term = pi/4.
Evaluating the remaining integral
Integral from 0 to 1 of x/(1 + x squared) dx.
Use the substitution u = 1 + x squared, du = 2x dx, so x dx = du/2.
When x = 0, u = 1. When x = 1, u = 2.
The integral becomes (1/2) times integral from 1 to 2 of du/u = (1/2)(ln 2 minus ln 1) = (1/2) ln 2.
Final answer
pi/4 minus (1/2) ln 2.
This is a clean, exam-typical result. The trick is recognising that arctan is the natural choice for u (because its derivative simplifies) and that the leftover integral needs its own substitution.
\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f\bigl(g(x)\bigr)\cdot g'(x)\int \frac{du}{1+u^2} = \arctan(u) + C\int \frac{du}{1+u} = \ln|1+u| + C\int u\,dv = uv - \int v\,duAll four of these should be on instant recall for Calculus II. The first is the FTC chain rule form. The second and third appear every time you do a u-substitution with an exponential denominator. The fourth is the integration by parts formula.
Students often forget to multiply by the derivative of the upper limit when using the FTC with a composite upper bound. If the upper limit is g(x), you must include g'(x) as a factor. Omitting it is one of the most common errors on exams.
Students sometimes confuse integral of 1/(1 + u squared) with integral of 1/(1 + u). The first gives arctan(u), the second gives ln|1 + u|. These are completely different results, and mixing them up costs full marks.
When using integration by parts, students occasionally pick u and dv the wrong way round (for example, setting u = dx and dv = arctan(x) dx). The LIATE rule is a useful mnemonic: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Functions earlier in this list are better choices for u.
When evaluating a definite integral after a u-substitution, students sometimes forget to change the limits of integration to match the new variable u. Either convert the limits, or back-substitute before evaluating.
The FTC chain rule problem is a near-guaranteed exam question. Expect a function defined as an integral with a non-trivial upper limit, and be asked for either f'(x) or the tangent line at a point.
U-substitution problems with e^x are extremely common on Calculus II midterms. You will almost certainly see at least one integral requiring u = e^x.
Integration by parts with arctan(x) is a classic definite integral problem. The combination of parts followed by a second substitution is a favourite exam structure.
Whenever you see 1/(1 + x squared) in an integrand, think arctan immediately.
True or False: d/dx of the integral from 0 to x squared of sin(t) dt equals sin(x squared). (Answer: False. You must multiply by the derivative of the upper limit, 2x.)
Fill in the blank: The integral of 1/(1 + u squared) du equals ________. (Answer: arctan(u) + C.)
True or False: When performing u-substitution on a definite integral, you may keep the original x-limits if you back-substitute before evaluating. (Answer: True.)
Fill in the blank: In integration by parts, the LIATE rule suggests choosing u as the function that comes ________ in the list. (Answer: earliest.)
True or False: The integral from 0 to 1 of arctan(x) dx = pi/4 minus ln(2)/2. (Answer: False. It equals pi/4 minus (1/2) ln 2, which is the same thing, so this one is actually True. Be careful with equivalent forms.)
Q: Let F(x) = the integral from 2 to (x cubed + 1) of ln(t) dt. Find F'(x).
A: F'(x) = ln(x cubed + 1) times 3x squared. Apply the FTC with the chain rule: evaluate the integrand at the upper limit, then multiply by the derivative of the upper limit.
Q: Evaluate the integral of e^x / (4 + e^(2x)) dx.
A: Set u = e^x, du = e^x dx. The integral becomes integral of du/(4 + u squared) = (1/2) arctan(u/2) + C = (1/2) arctan(e^x / 2) + C. This uses the general form: integral of 1/(a squared + u squared) du = (1/a) arctan(u/a) + C.
Q: Use integration by parts to evaluate the integral from 0 to 1 of x e^x dx.
A: Set u = x, dv = e^x dx. Then du = dx, v = e^x. The integral = x e^x from 0 to 1, minus the integral from 0 to 1 of e^x dx = (1 times e minus 0) minus (e minus 1) = e minus e + 1 = 1.
Q: Evaluate the integral of 1/(1 + e^x) dx. (Hint: multiply numerator and denominator by e^(negative x).)
A: Rewrite as integral of e^(negative x)/(e^(negative x) + 1) dx. Set u = e^(negative x) + 1, du = negative e^(negative x) dx. The integral becomes negative ln|e^(negative x) + 1| + C, which simplifies to x minus ln(1 + e^x) + C.
These techniques feed directly into the trigonometric integrals covered in Section 7.2 of the same course, where u-substitution and integration by parts are applied to products of sin and cos powers. The FTC chain rule reappears whenever an integral defines a function in later topics such as probability density functions and error functions. Integration by parts is also the foundation for deriving reduction formulas, which you will use extensively in later lectures.
Fundamental Theorem of Calculus, FTC Part 1, chain rule for integrals, Leibniz integral rule, u-substitution, change of variables, integration by parts, LIATE rule, arctan integral, inverse tangent, exponential integrals, definite integral evaluation, tangent line to integral function, Calculus II, NYU, Calculus for Engineers