MTH6131 Chap.15 Trigonometric Substitution and Partial Fractions
Trigonometric Substitution and Partial Fractions
Three substitutions for three roots
The schedule gives integration techniques three consecutive weeks, and the exercise sets split them into trigonometric substitution, partial fractions and further substitutions. A square root of a² − x² calls for x = a sin θ, a root of a² + x² for x = a tan θ, and a root of x² − a² for x = a sec θ.
In each case a Pythagorean identity turns the root into a single trigonometric function, and the integral becomes one the table can handle. The exercise solutions state the substitution, the differential and the transformed root in one sentence before anything else.
Back to x through a triangle
After integrating in θ, the answer has to be expressed in x again.
The substitution fixes two sides of a right triangle, Pythagoras gives the third, and every trigonometric ratio in the answer is read from it. When the quadratic under the root is not already a sum or difference of squares, completing the square first reduces it to one.
The integral of secant cubed appears often enough that it is derived once by parts and then quoted.
Partial fractions
A rational function whose denominator factors can be split into simpler fractions, each with a standard integral. If the numerator's degree is not below the denominator's, divide first.
Then write one term per factor: a constant over each distinct linear factor, a constant over each power of a repeated linear factor, and a linear numerator over each irreducible quadratic.
The constants come from substituting roots or comparing coefficients.
Further substitutions
The miscellaneous set adds substitutions that rationalise a root, such as u equal to the square root of x, and the half-angle substitution t = tan(x/2), which turns any rational function of sine and cosine into a rational function of t.
Different routes can give answers that look unrelated and differ only by a constant.
What this chapter covers
- 01
Choosing among sine, tangent and secant substitutions
- 02
Reference triangles for returning to x
- 03
Completing the square under a root
- 04
The integral of secant cubed
- 05
Dividing before decomposing
- 06
Partial fractions for linear, repeated and quadratic factors
- 07
Rationalising substitutions
- 08
The half-angle substitution
Worked example · free
A difference of squares, two ways
- 2Factor the denominator and set up the decomposition.
- 2Find the constants and integrate each term.
- 1Combine the logarithms.
Key terms
- Trigonometric Substitution
- Replacing x by a multiple of sine, tangent or secant so that a square root of a sum or difference of squares becomes a single trigonometric function.
- Reference Triangle
- A right triangle drawn from a substitution, used to express the trigonometric ratios in the answer in terms of x.
- Partial Fractions
- The decomposition of a rational function into a sum of simpler fractions, one for each factor of the denominator.
- Irreducible Quadratic
- A quadratic factor with no real roots, which takes a linear numerator in a partial fraction decomposition.
- Half-angle Substitution
- The substitution t equal to the tangent of half the angle, which turns rational functions of sine and cosine into rational functions of t.
- Improper Fraction
- A rational function whose numerator has degree at least that of its denominator, which must be divided before decomposing.
Trigonometric Substitution and Partial Fractions FAQ
How do I choose between sine, tangent and secant substitutions?
Read the root. A square root of a squared minus x squared takes x equals a sine theta, a root of a squared plus x squared takes a tangent, and a root of x squared minus a squared takes a secant. Each choice is the one whose Pythagorean identity turns the expression under the root into a perfect square.
What do I do when the degree of the numerator is too high for partial fractions?
Divide first. Polynomial long division turns the improper fraction into a polynomial plus a proper fraction, and only the proper fraction is decomposed. Trying to decompose an improper fraction directly produces equations for the constants that cannot be satisfied.
Two methods gave me answers that look different. Is one of them wrong?
Not necessarily. Antiderivatives of the same function can differ by a constant, and different substitutions often produce forms that look unrelated. Differentiate both answers: if each returns the integrand, both are correct, and their difference will turn out to be a constant.
Exam move
Draw the reference triangle for every trigonometric substitution you do, even when you could skip it, because the triangle is what makes the back-substitution reliable. For partial fractions, practise one example of each factor type until setting up the decomposition takes no thought.
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