npr=3024 find r

(2)dy/dx=secx-tanx/secx+tanx. plz help

1 answer

If npr=3024, r=3024/np

y' = (secx-tanx)/(secx+tanx)
y' = (secx-tanx)^2/tan^2x
y' = csc^2x - 2cscx + 1

Those are straightforward integrals
Similar Questions
  1. Evaluate the integral of(secx)^2 * (tanx)^3 *dx I started out with letting u=secx and du=secx*tanx*dx , but then I am kind of
    1. answers icon 1 answer
  2. tan(3x) + 1 = sec(3x)Thanks, pretend 3x equals x so tanx + 1 = secx we know the law that 1 + tanx = secx so tanx + 1 becomes
    1. answers icon 0 answers
  3. How do you verify:1.) sinxcosx+sinx^3secx=tanx 2.) (secx+tanx)/(secx-tan)=(secx+tanx)^2 I tried starting from the left on both
    1. answers icon 0 answers
  4. I can't find the integral for(tanx)^(6)*(secx)^(2) I tried splitting up tanx into (tanx)^2*(tanx)^4 and let the latter equal
    1. answers icon 0 answers
more similar questions