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Mathematical calculus! Find the inverse derivative (original function)
1 solution: because (tanx)'=cos2x(cos2x refers to the square of cosx), (x)'= 1.

And because f' = [1+cos2x]/cos2x =1/cos2x+1.

Get f=tanx+x+c(c is a constant).

2 solution: f' = cscxcottx = cosx/sin2x (sin2x refers to the square of sinx)

According to the knowledge of university calculus, FCSCX COT XDX = FCOSX/SIN2XDX (SIN2X refers to the square of sinx).

=f 1/sin2xd(sinx)

=-0.33 inch -3x+ Celsius

So f=-0.33sin-3x+ c(sin-3x refers to the negative cube of sinx, 0.33 finger 1/3).

It is difficult to express my meaning clearly because of the limitation of mathematical symbol input.

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