Prove that :\(\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a - x)\,dx\)Hence show that :\(\int_0^{\pi} \sin x\,dx = 2\int_0^{\pi/2} \sin x\,dx\)

Answer: Split the interval at \(a\) and substitute \(x = 2a - t\) in the second part; then apply it with \(f(x) = \sin x\), \(a = \frac{\pi}{2}\).

Proof: \(\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_a^{2a} f(x)\,dx\) ... (1)

In \(I_2 = \int_a^{2a} f(x)\,dx\), put \(x = 2a - t\), so \(dx = -dt\). When \(x = a\), \(t = a\); when \(x = 2a\), \(t = 0\).

\(I_2 = \int_a^0 f(2a - t)(-dt) = \int_0^a f(2a - t)\,dt = \int_0^a f(2a - x)\,dx\) (a definite integral does not depend on the variable).

Substituting in (1): \(\int_0^{2a} f(x)\,dx = \int_0^a f(x)\,dx + \int_0^a f(2a - x)\,dx\). Hence proved.

Hence part: Take \(f(x) = \sin x\) and \(2a = \pi\), i.e. \(a = \frac{\pi}{2}\).

\(\int_0^{\pi} \sin x\,dx = \int_0^{\pi/2} \sin x\,dx + \int_0^{\pi/2} \sin(\pi - x)\,dx = \int_0^{\pi/2} \sin x\,dx + \int_0^{\pi/2} \sin x\,dx = 2\int_0^{\pi/2} \sin x\,dx\).

(Check: both sides equal 2, since \(\int_0^{\pi}\sin x\,dx = [-\cos x]_0^{\pi} = 2\) and \(\int_0^{\pi/2}\sin x\,dx = 1\).)