Question

高等数学积分学一道简单的题若f'（sin^2 x）=cos^2 x ,则f（x）=?

Answer 1

[f((sinx)^2)]' = 2sinxcosx f'((sinx)^2)

=2sinx (cosx)^3

f( (sinx)^2 ) = 2 ∫sinx (cosx)^3 dx

= -(1/2)(cosx)^4 +C

=-(1/2)( 1- (sinx)^2 )^2 + C

f(x) =-(1/2)( 1- x )^2 + C