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• y = x² The slope at any point (x, y) is the function's derivative evaluated for the point (x, y); that is, y'(x) = 2x is the slope for any x value. You'll learn about the derivative as your course continues, if you haven't already. As you sai…

  in Help with parabola problem? Comment by siths-and-giggles June 2021
• y = x² The slope at any point (x, y) is the function's derivative evaluated for the point (x, y); that is, y'(x) = 2x is the slope for any x value. You'll learn about the derivative as your course continues, if you haven't already. As you sai…

  in Help with parabola problem? Comment by siths-and-giggles June 2021
• ∫ 1/((2 + cosx) sinx) dx Let t = tan(x/2) dt = 1/2 sec²(x/2) dx 2 cos²(x/2) dt = dx From tan(x/2) = t, you have sin(x/2) = t/√(1 + t²) cos(x/2) = 1/√(1 + t²) Squaring both sides, you have cos²(x/2) = 1/(1 + t²), so the equation contain…

  in How do ∫1/[(2+cosx)*sinx] dx? Comment by siths-and-giggles June 2021
• ∫ 1/((2 + cosx) sinx) dx Let t = tan(x/2) dt = 1/2 sec²(x/2) dx 2 cos²(x/2) dt = dx From tan(x/2) = t, you have sin(x/2) = t/√(1 + t²) cos(x/2) = 1/√(1 + t²) Squaring both sides, you have cos²(x/2) = 1/(1 + t²), so the equation contain…

  in How do ∫1/[(2+cosx)*sinx] dx? Comment by siths-and-giggles June 2021
• ∫ 1/((2 + cosx) sinx) dx Let t = tan(x/2) dt = 1/2 sec²(x/2) dx 2 cos²(x/2) dt = dx From tan(x/2) = t, you have sin(x/2) = t/√(1 + t²) cos(x/2) = 1/√(1 + t²) Squaring both sides, you have cos²(x/2) = 1/(1 + t²), so the equation contain…

  in How do ∫1/[(2+cosx)*sinx] dx? Comment by siths-and-giggles May 2021