Difficulty: Intermediate to Advanced | Prerequisites: Derivatives of inverse trig and hyperbolic functions, u-substitution, polynomial long division
This set of techniques covers three patterns that appear on the exam: evaluating definite integrals of hyperbolic functions (by rewriting in terms of exponentials), integrating expressions with sqrt(a^2 - x^2) (splitting, using inverse trig and u-substitution), and handling rational integrands where the degree of the numerator meets or exceeds that of the denominator (polynomial long division first, then standard forms).
cosh(x) (hyperbolic cosine)
Defined as (e^x + e^(-x)) / 2. Its derivative is sinh(x), and its integral is sinh(x) + C. Think of it as the "even" part of the exponential function.
sinh(x) (hyperbolic sine)
Defined as (e^x - e^(-x)) / 2. Its derivative is cosh(x), and its integral is cosh(x) + C. Think of it as the "odd" part of the exponential function.
Inverse trigonometric integrals
Standard results students must know: integral of 1/sqrt(a^2 - x^2) dx = arcsin(x/a) + C, and integral of 1/(a^2 + x^2) dx = (1/a) arctan(x/a) + C. These appear whenever the denominator has a sum or difference of squares.
Polynomial long division (for integrals)
When the degree of the numerator is greater than or equal to the degree of the denominator, divide first to get a polynomial plus a proper fraction. Then integrate each piece separately. In simple terms, you "simplify" the fraction before integrating, the same way you might simplify 7/3 into 2 + 1/3.
u-substitution
Replacing part of the integrand with a new variable to simplify the integral. The substitution u = g(x) transforms the integral into one in terms of u.
cosh(x) = (e^x + e^(-x)) / 2 and sinh(x) = (e^x - e^(-x)) / 2.
The integral of cosh(x) is sinh(x) + C. The integral of sinh(x) is cosh(x) + C.
To evaluate a definite integral of cosh or sinh, you can either use the sinh/cosh antiderivative directly or convert to exponentials. The "Show Your Work" version of the exam expects the exponential conversion.
Evaluate integral from 0 to ln(5) of cosh(x) dx.
Step 1: Antiderivative of cosh(x) is sinh(x).
sinh(x) evaluated from 0 to ln(5)
Step 2 (showing work via exponentials):
sinh(x) = (e^x - e^(-x)) / 2
At x = ln(5): (e^(ln 5) - e^(-ln 5)) / 2 = (5 - 1/5) / 2 = (25 - 1) / 10 = 24/10 = 12/5
At x = 0: (e^0 - e^0) / 2 = 0
Result: 12/5
The key insight is that e^(ln a) = a and e^(-ln a) = 1/a. These identities turn the hyperbolic evaluation into straightforward arithmetic.
When you see a sum in the numerator over sqrt(a^2 - x^2), split it into separate integrals and handle each with the appropriate technique.
Evaluate integral of (6 + x) / sqrt(36 - x^2) dx.
Split into two integrals:
I = 6 * integral of 1/sqrt(36 - x^2) dx + integral of x/sqrt(36 - x^2) dx
First integral (inverse trig form):
integral of 1/sqrt(36 - x^2) dx = arcsin(x/6) + C1
So the first part = 6 arcsin(x/6) + C1
This uses the standard form: integral of 1/sqrt(a^2 - x^2) dx = arcsin(x/a) + C, with a = 6.
Second integral (u-substitution):
Let u = 36 - x^2, then du = -2x dx, so x dx = -du/2
integral of x/sqrt(36 - x^2) dx = -(1/2) integral of u^(-1/2) du = -sqrt(u) + C2 = -sqrt(36 - x^2) + C2
Combining:
I = 6 arcsin(x/6) - sqrt(36 - x^2) + C
When the degree of the numerator is greater than or equal to the degree of the denominator, you cannot integrate the fraction as it stands. Divide first, then integrate the quotient and the remainder separately.
Evaluate integral of (x^4 - 7) / (9 + x^2) dx.
The numerator has degree 4, the denominator has degree 2, so perform polynomial long division:
x^4 divided by (x^2 + 9):
x^4 / (x^2 + 9) = x^2 - 9 + 81/(x^2 + 9)
So the integrand becomes: (x^4 - 7)/(x^2 + 9) = x^2 - 9 + 81/(x^2 + 9) - 7/(x^2 + 9) = x^2 - 9 + 74/(x^2 + 9)
Wait, let us redo this carefully. x^4 - 7 divided by x^2 + 9:
x^4 / (x^2 + 9): quotient starts with x^2. x^2 * (x^2 + 9) = x^4 + 9x^2. Subtract: (x^4 - 7) - (x^4 + 9x^2) = -9x^2 - 7.
Next term: -9x^2 / (x^2 + 9): quotient term -9. -9 * (x^2 + 9) = -9x^2 - 81. Subtract: (-9x^2 - 7) - (-9x^2 - 81) = 74.
Result: (x^4 - 7)/(x^2 + 9) = x^2 - 9 + 74/(x^2 + 9)
Now integrate term by term:
integral of x^2 dx = x^3/3
integral of -9 dx = -9x
integral of 74/(x^2 + 9) dx: use the standard form integral of 1/(a^2 + x^2) dx = (1/a) arctan(x/a). Here a = 3. So this gives 74 * (1/3) arctan(x/3) = (74/3) arctan(x/3).
Final answer: x^3/3 - 9x + (74/3) arctan(x/3) + C
Students often try to integrate (x^4 - 7)/(x^2 + 9) directly without dividing first. If the numerator's degree is at least as large as the denominator's, you must do polynomial long division before integrating.
When evaluating sinh(ln a), students sometimes forget that e^(-ln a) = 1/a, not -a. The negative is in the exponent, not in the base.
Students confuse the two standard inverse trig forms: 1/sqrt(a^2 - x^2) gives arcsin, while 1/(a^2 + x^2) gives arctan. Mixing these up is an easy way to lose all the marks on a problem.
In Problem 17A, students sometimes try to do the entire integral with one substitution. Splitting the fraction into two separate integrals (one for the constant, one for the x term) and handling each with its own technique is the intended approach.
Hyperbolic function integrals are tested on this exam. Know cosh and sinh in terms of exponentials and be ready to evaluate sinh(ln a) by converting to (a - 1/a)/2.
The "separate and conquer" strategy for integrands like (6 + x)/sqrt(36 - x^2) appears regularly. Recognise the pattern: constant over sqrt(a^2 - x^2) calls for arcsin; x over sqrt(a^2 - x^2) calls for u-substitution.
Polynomial long division is a prerequisite step, not an integration technique in itself. If you skip the division and try partial fractions or substitution on an improper rational function, you will get stuck.
On the "Show Your Work" portion of this exam, writing out the long division and the substitution steps earns partial credit even if the final arithmetic has an error.
Fill in the blank: cosh(x) = (e^x + ____) / 2. (Answer: e^(-x))
True or false: The integral of cosh(x) is -sinh(x) + C. (False. It is sinh(x) + C, no negative sign.)
Fill in the blank: integral of 1/sqrt(25 - x^2) dx = arcsin(____) + C. (Answer: x/5)
True or false: To integrate x^5 / (x^2 + 1) dx, you should do polynomial long division first. (True. The numerator has higher degree than the denominator.)
Fill in the blank: integral of 1/(16 + x^2) dx = (1/) arctan(x/) + C. (Answer: 4 and 4)
Q: Evaluate integral from 0 to ln(3) of sinh(x) dx.
A: Antiderivative of sinh(x) is cosh(x). cosh(ln 3) = (e^(ln 3) + e^(-ln 3))/2 = (3 + 1/3)/2 = (10/3)/2 = 5/3. cosh(0) = (1 + 1)/2 = 1. Result: 5/3 - 1 = 2/3.
Q: Evaluate integral of (3 + x)/sqrt(49 - x^2) dx.
A: Split into 3 * integral of 1/sqrt(49 - x^2) dx + integral of x/sqrt(49 - x^2) dx. First part: 3 arcsin(x/7). Second part: let u = 49 - x^2, du = -2x dx, giving -sqrt(49 - x^2). Answer: 3 arcsin(x/7) - sqrt(49 - x^2) + C.
Q: Evaluate integral of (x^3 + 2)/(x^2 + 4) dx.
A: Degree of numerator (3) >= degree of denominator (2), so divide: x^3/(x^2 + 4) = x - 4x/(x^2 + 4). So the integrand is x - 4x/(x^2 + 4) + 2/(x^2 + 4). Integrate: x^2/2 - 2 ln|x^2 + 4| + arctan(x/2) + C. (The 4x/(x^2+4) piece uses u = x^2 + 4; the 2/(x^2+4) piece uses the arctan form with a = 2.)
Q: What is the value of e^(-ln 7)?
A: e^(-ln 7) = 1/e^(ln 7) = 1/7. This identity is essential for evaluating hyperbolic functions at x = ln(a).
These integration techniques feed directly into the application problems elsewhere in the course. Arc length integrals sometimes require completing the square or trig substitution. Volumes of revolution produce integrands that need polynomial expansion or u-substitution. Later in the course, techniques of integration (integration by parts, trig integrals, partial fractions) build on the same pattern-recognition skills practised here.
Related Terms / Search Tags
hyperbolic cosine, hyperbolic sine, cosh, sinh, definite integral, inverse trig integrals, arcsin, arctan, polynomial long division, u-substitution, sqrt(a^2 - x^2), improper rational function, integration techniques, M 408D, 2414 exam, UT Austin calculus, integral calculus techniques