f'(x) = cos/(sinx)^3

We know that :

f(x) = integra; f'(x)

==> f(x) = intg (cosx/(sinx)^3 dx

Let t = sinx

==> dt = cosdx

==> f(x) = intg (dt/t^3)

==> f(x) = intg (t^-3) dt

= t^-2/-2 + C

= -1/2t^2 + C

Now substitute with t= sinx

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

f'(x) = cosx/(sinx)^3.

Therefore f(x) = Int (cosxdx/(sinx)^3}....(1)

We put sinx = t. then cosxdx = dt. So with this transformation the cosxdx(/sinx)^3 in (1) becomes:

f(x) = Int {dt/t^3}

f(x) = (1/(-3+2))t^(3+1) +C = (-1/2) /t^2 = (-1/2)(sinx)^2 + const.

f(x) = -1/(2(sinx)^2) +const.

To determine the function F(x) = f(x), we'll have to integrate cosx/(sinx)^3.

Int cosxdx/(sinx)^3 = F(x) + C

To determine the integral of f'(x), we'll change the variable x.

We'll note sin x = t

We'll differentiate both sides:

cos xdx = dt

We'll substitute sin x by t and cosxdx by dt and we'll get:

Int cosxdx/(sinx)^3 = Int dt/t^3

We'll use the property of the negative exponent:

1/t^3 = t^-3

Int dt/t^3 = Int t^-3dt

Int t^-3dt = t^(-3+1)/(-3+1) + C

Int t^-3dt = t^-2/-2 + C

Int t^-3dt = -1/2t^2 + C

But t = sin x

**Int cosxdx/(sinx)^3 = 1/2(sin x)^2 + C**

**So, F(x) = 1/2(sin x)^2 + C.**