posted by Rory .
Let f(t) be an odd function
Using properties of the deﬁnite integral plus simple
substitution show that if f is continuous on [−a, a] for a positive number a, then
f(t) dt = 0
The definition of an odd function is
If f(x) is continuous on the interval [-a,a], then
I=∫f(t)dt from -a to a
= I1 + I2
I2 = ∫f(t)dt from 0 to a
I1 = ∫f(t)dt from -a to 0
Using property of odd function, f(-x)=-f(-x), and substitute u=-t,dt=-du
I1 = ∫-f(u)(-du) from a to 0
=-∫f(u)du from 0 to a
Incidentally, as exercise, you can use the above fact to solve:
∫[x^3 + x^4(tan^3(x))]dx from -π/4 to π/4.