Question

щий вид первообразных и: 1 а) f(x) = 1/ 4√x -x^2б) f(x) = 2 cos x- 1 /sin2 x ​

Answer 1

∫f(x)dx = ∫1/4√x -x^2 dx= 1/6 x^(3/2)-1/3x^3

∫f(x) = ∫2 cos x- 1 /sin2 x ​dx=2sinx-∫dx/sin2x=

=2sinx-1/2∫cosxdx/sinxcos^2x=2sinx-1/2∫dt/t(1-t^2)=

sinx=t

=2sinx-1/2∫dt/t-1/2∫tdt/(1-t^2)=2sinx-(1/2ln|sinx|-1/4(ln(cos^2x))