Definite Integration Techniques

Consider this integral:

If you try to expand \sin^5 x and integrate term by term, you are in for a long afternoon. But notice: \cos x is the derivative of \sin x. That single observation turns the entire integral into something you can do in your head.

Set u = \sin x. Then du = \cos x \, dx — and the integrand is just u^5 \, du. The antiderivative is u^6/6. Plug in the limits (converting them too: when x = 0, u = 0; when x = \pi/2, u = 1) and you get 1/6 - 0 = 1/6.

The whole problem, from start to answer, takes four lines. The trick was recognising a substitution and knowing how to carry the limits along with it.

That is what this article is about: three systematic techniques — substitution, integration by parts, and special substitutions — that convert difficult definite integrals into tractable ones. You already know the indefinite versions of these techniques. The definite-integral versions add one critical rule: when you change the variable, you must change the limits too.

Substitution in definite integrals

The idea

Substitution in an indefinite integral says: if the integrand looks like f(g(x)) \cdot g'(x), set u = g(x) and write du = g'(x) \, dx. The integral becomes \int f(u) \, du, which is simpler.

For a definite integral, the same logic applies — but you also need to convert the limits of integration. If the original integral runs from x = a to x = b, and you set u = g(x), then the new limits are u = g(a) to u = g(b).

The key point: once you switch to u, you never go back to x. You evaluate the antiderivative at u = g(b) and u = g(a), and you are done. There is no need to undo the substitution.

Why the limits must change

The original integral computes a specific number — the signed area under a curve from x = a to x = b. When you change the variable to u, you are looking at a different curve (the transformed one), and the interval over which you integrate must correspond to the same region. If x runs from 0 to \pi/2, and u = \sin x, then u runs from \sin 0 = 0 to \sin(\pi/2) = 1. Keeping the old limits would compute the area over the wrong interval.

A subtlety: limits can flip

Sometimes g(a) > g(b), so the new lower limit is larger than the new upper limit. That is fine — it just means the integral picks up a sign change, which is correct.

For example, in \int_0^\pi \cos x \cdot (-\sin x) \, dx, if you set u = \cos x, then u(0) = 1 and u(\pi) = -1, so you integrate from 1 to -1. You can either leave it as \int_1^{-1} u \, du and evaluate (getting the correct sign automatically), or flip the limits and add a minus sign: -\int_{-1}^{1} u \, du.

Integration by parts in definite integrals

The product rule, run backward

Integration by parts comes from the product rule. If u(x) and v(x) are differentiable functions, then

Integrate both sides from a to b:

The left side, by the fundamental theorem, is just [u \, v]_a^b = u(b)v(b) - u(a)v(a). Rearranging:

The definite form differs from the indefinite one in exactly one way: there is a boundary term [uv]_a^b that you must evaluate. In an indefinite integral, this boundary term does not appear — it is absorbed into the constant of integration.

Choosing u and v'

The classic guideline is LIATE: when the integrand is a product of two types of function, let u be whichever type appears first in the list:

Logarithmic — Inverse trigonometric — Algebraic — Trigonometric — Exponential

This is a heuristic, not a law. It works because functions earlier in the list tend to simplify when differentiated (logarithms become algebraic, algebraic expressions drop in degree), while functions later in the list do not get worse when integrated.

Repeated by-parts and reduction formulas

Some integrals require applying by-parts more than once. A common pattern is \int_0^{\pi/2} \sin^n x \, dx, where applying by-parts once reduces the power from n to n - 2:

This is a reduction formula — it expresses the integral with index n in terms of the integral with a smaller index. Applying it repeatedly, you reduce all the way down to \int_0^{\pi/2} \sin^0 x \, dx = \pi/2 (when n is even) or \int_0^{\pi/2} \sin^1 x \, dx = 1 (when n is odd). The result is the Wallis formula:

where n!! = n(n-2)(n-4)\cdots is the double factorial.

Deriving the reduction formula

Start with I_n = \int_0^{\pi/2} \sin^n x \, dx. Write \sin^n x = \sin^{n-1} x \cdot \sin x. Set u = \sin^{n-1} x and v' = \sin x, so u' = (n-1)\sin^{n-2} x \cos x and v = -\cos x. By parts:

The boundary term vanishes: at x = \pi/2, \cos(\pi/2) = 0; at x = 0, \sin^{n-1}(0) = 0. Replace \cos^2 x with 1 - \sin^2 x:

Bring (n-1)I_n to the left: I_n + (n-1)I_n = (n-1)I_{n-2}, so nI_n = (n-1)I_{n-2}, giving I_n = \frac{n-1}{n} I_{n-2}.

Typical substitutions

Some integrands do not obviously factor as f(g(x)) \cdot g'(x), but a well-chosen substitution transforms them into standard forms. Here are the three families of substitution that appear most often.

Trigonometric substitutions

When the integrand contains \sqrt{a^2 - x^2}, \sqrt{a^2 + x^2}, or \sqrt{x^2 - a^2}, a trigonometric substitution eliminates the square root.

When applying these to definite integrals, convert the limits to values of \theta. For instance, if x runs from 0 to a and you set x = a\sin\theta, then \theta runs from 0 to \pi/2.

The Weierstrass substitution

For integrals involving rational functions of \sin x and \cos x, the substitution t = \tan(x/2) converts everything into a rational function of t. The key identities:

This is a universal substitution — it always works for rational trigonometric integrands — but it can make the algebra heavy. Use it as a last resort when symmetry-based tricks do not apply.

The substitution x = a + b - t

For definite integrals over [a, b], the substitution x = a + b - t replaces x with its "mirror image" across the midpoint of the interval. This transforms \int_a^b f(x)\, dx into \int_a^b f(a + b - t)\, dt. When f(x) + f(a + b - x) simplifies (for example, equals a constant), adding the original integral to the transformed one gives twice the answer, and you can solve immediately.

This is not a new technique — it is the substitution version of the King property of definite integrals. But recognising when to apply it is a skill in itself.

Worked examples

Common confusions

• Forgetting to change the limits when substituting. This is the single most common error. If you set u = g(x), the new limits are g(a) and g(b), not a and b. If you forget, you will compute the antiderivative correctly but evaluate it at the wrong points, getting a wrong answer.

• Changing the limits and then also back-substituting. Pick one approach. Either change the limits (and stay in the u-world) or keep the old limits (and convert back to x before evaluating). Doing both — changing limits and also replacing u with g(x) — undoes the work and produces nonsense.

• Dropping the boundary term in by-parts. In an indefinite integral, the uv term is absorbed into the answer. In a definite integral, you must evaluate [uv]_a^b and include it. The boundary term is often the entire reason the answer comes out clean.

• Using a trigonometric substitution when a simpler one works. The presence of \sqrt{a^2 - x^2} does not automatically mean you need x = a\sin\theta. If there is also an x \, dx factor, u = a^2 - x^2 is simpler and faster. Try the obvious substitution first.

• Mishandling the sign when limits flip. If after substitution the upper limit is smaller than the lower limit, do not panic. Either evaluate directly (the minus sign takes care of itself), or flip the limits and explicitly insert a negative sign: \int_b^a = -\int_a^b.

Going deeper

Why the Weierstrass substitution always works

Any rational function of \sin x and \cos x — that is, any expression built from \sin x, \cos x, and the four arithmetic operations — can be written as a ratio of two polynomials in \sin x and \cos x. The identities

follow from the double-angle formulas \sin x = 2\sin(x/2)\cos(x/2) and \cos x = \cos^2(x/2) - \sin^2(x/2), after dividing numerator and denominator by \cos^2(x/2). With t = \tan(x/2), both \sin x and \cos x become rational functions of t, and dx = \frac{2}{1+t^2}\,dt is also rational in t. So the entire integrand becomes a rational function of t, which you can always decompose by partial fractions.

This is a powerful existence result: it guarantees that every integral of a rational trigonometric function can be reduced to partial fractions, even if the algebra is unpleasant.

Reduction formulas for other powers

The reduction formula I_n = \frac{n-1}{n} I_{n-2} derived above for \int_0^{\pi/2} \sin^n x \, dx has analogues for other families:

• \int_0^{\pi/2} \cos^n x \, dx = \frac{n-1}{n} \int_0^{\pi/2} \cos^{n-2} x \, dx — identical to the sine version, because \sin and \cos are interchangeable on [0, \pi/2].

• \int_0^{\pi/4} \tan^n x \, dx = \frac{1}{n-1} - \int_0^{\pi/4} \tan^{n-2} x \, dx — this one alternates in sign as you reduce.

• \int_0^1 x^n e^x \, dx = e - n\int_0^1 x^{n-1} e^x \, dx — from integrating x^n e^x by parts, with u = x^n and v' = e^x.

Each of these can be iterated to reduce the index all the way to a base case, producing a closed-form answer.

Recognising hidden substitutions

Many competition-level definite integrals are designed so that a single substitution collapses the problem, but the substitution is not obvious. A useful habit: when the integrand involves f(x) and f(a + b - x) (or f(a - x) when one limit is 0), adding the integral to its transformed version often simplifies. This is a technique, not a formula — the art is in spotting the symmetry.

For example, \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx. Set x = \pi/2 - t, so \sin x = \cos t and \cos x = \sin t. The integral becomes \int_0^{\pi/2} \frac{\cos t}{\cos t + \sin t}\,dt. Add the two versions:

So I = \pi/4. A problem that looks intractable becomes trivial once you notice the symmetry.

Where this leads next

• Leibniz Rule — how to differentiate an integral whose limits are themselves functions of a parameter.
• Definite Integrals — Inequalities — bounding and estimating integrals when you cannot (or do not need to) compute them exactly.
• Improper Integrals — what happens when one of the limits is infinity, or the integrand blows up inside the interval.
• Area Under Curve — the geometric application that makes all these techniques useful: computing actual areas, volumes, and physical quantities.
• Sum of Series Using Integration — converting infinite sums into definite integrals and evaluating them with these same tools.