Đổi biến \(x = \frac{\pi }{2} - t\), ta có
\(\int \limits_0^{\frac{\pi }{2}} f\left( {\sin x} \right)dx = - \int \limits_{\frac{\pi }{2}}^0 f\left[ {\sin \left( {\frac{\pi }{2} - x} \right)} \right]dx = \int \limits_0^{\frac{\pi }{2}} f\left( {\cos x} \right)dx\)
Vậy \(\int \limits_0^{\frac{\pi }{2}} f\left( {\sin x} \right)dx = \int \limits_0^{\frac{\pi }{2}} f\left( {\cos x} \right)dx\)
-- Mod Toán 12