Apr 03, 18 · The answer is =ln(tanxsecx)sinxC "Reminder" intsecxdx=ln(tanxsecx)C Therefore, the integral is int(tan^2xdx)/(secx)= intcosxtan^2xdx =intcosx(sec^2x1)dx =int(secxcosx)dx =ln(tanxsecx)sinxCAs x varies, the point (cos x, sin x) winds repeatedly around the unit circle centered at (0, 0) The point (, )goes only once around the circle as t goes from −∞ to ∞, and never reaches the point (−1, 0), which is approached as a limit as t approaches ±∞ As t goes from −∞ to −1, the point determined by t goes through the part of the circle in the third quadrant, fromIntegral CSC^2 3x cot 3x dx integral sec^4 (1 x) tan (1 x) dx;
Derivatives Of Trigonometric Functions
