令a=arcsin5x;a∈[-π/2,π/2]
则有sina=5x
cos(arcsin5x)=cosa=±√(1-sin^2(a))=±√(1-25x^2)
又∵a∈[-π/2,π/2]
∴cosa≥0
所以原式=√(1-25x^2)
且x∈[-1/5,1/5]