Concepts: substitution, trigonometric identities and partial fractions

Integrals22 min readLesson 3 of 8

3.1 Substitution Substitution is the chain rule read backwards. When the integrand contains a function and its derivative, name the inner function u and replace its differential du for the matching part of the integrand. The integral then becomes a standard form in u, which is integrated and converted back. For a definite integral there are two valid routes: change the limits into u values and never convert back, or convert back to x and use the original limits. Mixing the two routes gives nonsense, and examiners see it often. The skill is in recognising the pair. In the integral of two x cos of x squared, the inner function is x squared and its derivative is two x, so the substitution is immediate. In the integral of tangent, the integrand is sine over cosine, and the substitution is the cosine itself, which turns the integral into a logarithm. When a linear expression sits inside a power, the substitution is that linear expression and its derivative is a constant.

\int f (g (x)) g^{'} (x) d x = \int f (u) d u with u = g (x)

The substitution rule

\int tan x d x = - lo g ∣ cos x ∣ + C = lo g ∣ sec x ∣ + C

Tangent by substituting the cosine

3.2 Trigonometric identities Products and powers of sine and cosine are reduced with identities before integration. The square of sine is half of one minus cosine of twice the angle, and the square of cosine is half of one plus that cosine. Their product is half of sine of twice the angle. For odd powers, keep one factor for the substitution and convert the rest with the Pythagorean identity; for even powers, use double angle identities until the powers fall. The half angle results are what make integrals such as the sine squared of x manageable. Example: sine squared. The integral of sine squared from zero to pi is the integral of half of one minus cosine two x, which is half of the interval length, so the answer is pi by two. The same identity handles every even power after enough applications.

sin^{2} x = \frac{1 - cos 2 x}{2}, cos^{2} x = \frac{1 + cos 2 x}{2}, sin x cos x = \frac{sin 2 x}{2}

The three identities that remove powers

3.3 Partial fractions A rational integrand is split into simpler fractions whose denominators are the factors of the original denominator. For distinct linear factors the constants are found by substituting the roots. For a repeated factor a term is needed for each power of that factor. For an irreducible quadratic factor the numerator is linear. Once split, each piece is a standard logarithm or inverse tangent. The method requires the numerator degree to be lower than the denominator degree; if it is not, divide first. The three shapes that appear most often are one over x minus a, which integrates to a logarithm, one over a repeated linear factor, which integrates to a reciprocal, and a linear numerator over an irreducible quadratic, which splits into a logarithm plus an inverse tangent.

\frac{p x + q}{( x - a ) ( x - b )} = \frac{A}{x - a} + \frac{B}{x - b}

Distinct linear factors

\frac{p x + q}{( x - a ) ^{2}} = \frac{A}{x - a} + \frac{B}{( x - a ) ^{2}}

A repeated linear factor

\frac{p x + q}{x ^{2} + b x + c} = \frac{A ( 2 x + b )}{x ^{2} + b x + c} + \frac{B}{x ^{2} + b x + c}

Irreducible quadratic: split the numerator to create the derivative first

3.4 Worked examples Example 1: substitution. Integrate x times e to the power x squared. Put u equal to x squared, then du is two x dx, so the integral becomes half the integral of e to the u du, which is half e to the power x squared plus C. Example 2: identity. Integrate sine cubed x. Write it as sine x times one minus cosine squared x, put u equal to cosine x, and the integral becomes minus the integral of one minus u squared du, which is minus cosine x plus cosine cubed x over three plus C. Example 3: partial fractions. Integrate one over x squared minus one. The denominator factors as (x minus one)(x plus one), and the split is half over (x minus one) minus half over (x plus one). The integral is half the logarithm of the ratio of x minus one to x plus one, plus C.

\int x e^{x^{2}} d x = \frac{1}{2} e^{x^{2}} + C

Substitution u = x squared

\int \frac{d x}{x ^{2} - 1} = \frac{1}{2} lo g \frac{x - 1}{x + 1} + C

Partial fractions on a difference of squares

3.5 Common mistakes and exam tip Forgetting to replace dx by du and leaving the old differential in the integral is the first classic error. A second is choosing a substitution whose derivative is not present and forcing it anyway. In partial fractions, an incomplete split, such as using only one constant for a repeated factor, gives an answer that cannot be checked. The quick check is to differentiate the final answer; it should return the original integrand.

3.6 A repeated factor in partial fractions When the denominator has a factor squared, the split needs two terms for that factor. For the integrand one over x plus one whole squared times x minus one, write A over x plus one, B over x plus one whole squared and C over x minus one, then clear the denominator and compare coefficients or substitute the convenient values. The first term integrates to a logarithm, the second to a reciprocal and the third to a logarithm. Leaving out the B term is the standard error, and the coefficient comparison detects it immediately because the numerators will not match.

3.7 When to try identities before substitution If the integrand contains a product of sines and cosines, try an identity first, because the product often collapses to a single trigonometric function of a doubled angle. If the integrand contains a power of tangent or secant, keep one factor for the differential and convert the rest with the Pythagorean identity. Only when neither works should a general substitution be attempted, because a forced substitution usually makes the integral longer.