When the easy methods run out
You already own two great tools. u-substitution reverses the chain rule — it works when the integrand hides an inner function whose derivative is also sitting there. Integration by parts reverses the product rule — it shines when you can trade a hard integral for an easier one. But some integrands answer to neither. Two of the most common stumbling blocks are a stubborn square root like sqrt(a^2 - x^2), and a ratio of polynomials like (3x + 5) / (x^2 - x - 2). This guide hands you the standard tactic for each.
The common thread is the same one you have always relied on: change the problem into a shape you recognize. A plain u-substitution swaps one variable for another and hopes the new integral is friendlier. Here we are bolder — we substitute a whole trig function, or we tear a fraction apart — but the spirit is identical. Transform, integrate, then translate back. Keep that three-beat rhythm in mind and both methods feel less like magic and more like bookkeeping.
Trig substitution: three roots, three keys
Trigonometric substitution is built on three Pythagorean identities, each one a key cut to fit a particular root. Because 1 - sin^2(theta) = cos^2(theta), setting x = a sin(theta) turns sqrt(a^2 - x^2) into a cos(theta) — the root vanishes into a clean cosine. Because 1 + tan^2(theta) = sec^2(theta), setting x = a tan(theta) turns sqrt(a^2 + x^2) into a sec(theta). And because sec^2(theta) - 1 = tan^2(theta), setting x = a sec(theta) turns sqrt(x^2 - a^2) into a tan(theta). One look at the root tells you which substitution to grab.
sqrt(a^2 - x^2) set x = a*sin(theta) -> root = a*cos(theta) sqrt(a^2 + x^2) set x = a*tan(theta) -> root = a*sec(theta) sqrt(x^2 - a^2) set x = a*sec(theta) -> root = a*tan(theta)
There is one more piece of bookkeeping: when you replace x you must also replace dx by its differential. For x = a sin(theta) that means dx = a cos(theta) d(theta). Substitute both, and the entire integral becomes a problem in theta — usually a manageable mix of sines, cosines, or secants, which you then attack with the trig-power techniques in the next guide on powers of trig functions.
Walking through one — and the sign you must watch
Picture the integral of 1 / sqrt(4 - x^2) dx. The root sqrt(4 - x^2) matches the first shape with a = 2, so set x = 2 sin(theta), giving dx = 2 cos(theta) d(theta) and root = 2 cos(theta). The integral collapses to (2 cos(theta) d(theta)) / (2 cos(theta)), which is simply the integral of d(theta) — equal to theta. Now translate back. Since x = 2 sin(theta), we have sin(theta) = x/2, so theta = arcsin(x/2). The answer is arcsin(x/2) + C, an inverse function appearing right where the geometry promised it would.
To translate a messier answer back to x, draw the right triangle the substitution encodes. From sin(theta) = x/2, label the opposite side x and the hypotenuse 2; the Pythagorean theorem fills in the adjacent side as sqrt(4 - x^2). Then any trig function of theta you need — cos(theta), tan(theta), sec(theta) — you simply read off the triangle. This little picture is the bridge that carries every theta-answer back into the original x.
Partial fractions: splitting a rational function
Now the second wall: a ratio of polynomials. Partial fraction decomposition is the exact reverse of adding fractions over a common denominator. You take one tangled fraction and split it back into a sum of simpler ones, each with a basic factor of the original denominator on the bottom — and each of those small pieces is something you already know how to integrate. This is the engine behind integrating rational functions in general.
- Check the degrees. If the numerator's degree is not lower than the denominator's, do polynomial long division first and set the polynomial quotient aside (you integrate it trivially).
- Factor the denominator completely into linear factors (x - a) and irreducible quadratic factors (x^2 + bx + c).
- Write one term per factor: a linear (x - a) gives A/(x - a); a repeated (x - a)^2 also needs B/(x - a)^2; an irreducible quadratic gets a linear top, (Bx + C)/(x^2 + bx + c).
- Solve for the unknown constants — clear denominators and either match coefficients or plug in convenient x-values — then integrate each piece into a logarithm, an arctangent, or a power.
Take (3x + 5) / (x^2 - x - 2). The bottom factors as (x - 2)(x + 1), so we write it as A/(x - 2) + B/(x + 1). Clearing denominators gives 3x + 5 = A(x + 1) + B(x - 2). Plug in x = 2: 11 = 3A, so A = 11/3. Plug in x = -1: 2 = -3B, so B = -2/3. The integral is now just (11/3) ln|x - 2| - (2/3) ln|x + 1| + C — two ordinary logs, no cleverness required.
Choosing your weapon — and an honest limit
The decision is mostly visual. A square root of a quadratic shape (a^2 ± x^2 or x^2 - a^2) says trig substitution. A ratio of polynomials with no root says partial fractions. The two even cooperate: when a quadratic in a root is not centered, complete the square first — sqrt(x^2 + 4x + 13) becomes sqrt((x + 2)^2 + 9), and a shift u = x + 2 hands you the clean a^2 + u^2 shape ready for a tangent substitution.
There is a remarkable promise hiding here. Because every partial-fraction piece integrates to a log, an arctangent, or a power, every rational function has an elementary antiderivative — there are no exceptions and nothing exotic is ever needed. Partial fractions also resurface later: the inverse Laplace transform leans on exactly this split, breaking a transfer function into simple terms you read off one by one. And for some integrands of trig fractions, the Weierstrass substitution t = tan(theta/2) turns the whole thing into a rational function of t — handing it straight to partial fractions.