Properties of definite integrals

P0: the letter used for the variable does not matter, ∫ₐᵇ f(x) dx = ∫ₐᵇ f(t) dt.

P1: swapping the limits changes the sign, ∫ₐᵇ f dx = −∫ᵇₐ f dx; in particular ∫ₐᵃ f dx = 0.

P2: an interval can be split at any c, ∫ₐᵇ f dx = ∫ₐᶜ f dx + ∫ᶜᵇ f dx. This is how a modulus or a piecewise function is handled.

P3: ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx, and its special case P4: ∫₀ᵃ f(x) dx = ∫₀ᵃ f(a − x) dx. Adding the original integral to the reflected one often cancels the hard part.

P5: an integral over [0, 2a] splits as ∫₀ᵃ f(x) dx plus ∫₀ᵃ f(2a − x) dx. P6: so when f(2a − x) = f(x) the whole is twice ∫₀ᵃ f(x) dx, and when f(2a − x) = −f(x) it is 0.

P7: on a symmetric interval, ∫₋ₐᵃ f dx = 2∫₀ᵃ f dx if f is even, and 0 if f is odd.

Worked example (P2): x³ − x changes sign at −1, 0 and 1, so ∫₋₁² |x³ − x| dx is split into three pieces: 1/4 + 1/4 + 9/4 = 11/4.

Worked example (P4): with I = ∫₀^(π/2) sin⁴x/(sin⁴x + cos⁴x) dx, replacing x by π/2 − x swaps sin and cos. Adding the two forms gives 2I = π/2, so I = π/4.

Worked example (P3): I = ∫ from π/6 to π/3 of dx/(1 + √tan x); replacing x by π/2 − x and adding gives 2I = π/6, so I = π/12.

Worked example (P7): sin⁵x cos⁴x is odd, so its integral from −1 to 1 is 0 without any working. sin²x is even, so ∫ from −π/4 to π/4 of sin²x dx = 2∫ from 0 to π/4 = π/4 − 1/2.

Using P4 and P6 together, ∫₀^(π/2) log sin x dx = −(π/2) log 2.