Recent content by Krash

Sorry, x = -sinht, because the domain of the integral is negative. i switched around the dx and x substitutions the second time around.

separating into horizontal and vertical components would be a good idea. If you consider just the vertical part of the problem, the period of the jump is going to just be how long it takes gravity to overcome the skater's upward velocity.

this is the current solution I am working on: dx = cosht dt x = -sinht therefore; 1/ sqrt(x^2 - 1 ) = cosht dt / sqrt( -sinh^2 t - 1 ) ----- (1) using the trig id: cosh^2 t - sinh^2 t = 1 -sinh^2 t = -cosh^2 t + 1 ----- (2) (2) into (1) int (cosht / sqrt(-cosh^2 t)) dt =...

cheers. that still means i can't sub cosht for x but i can use sinh = x then everything cancels and I am left with int dt which is -sinh^-1(x) which gives me: (1 / (-sinh(-3))) - (1 / (-sinh(-2))) = -0.175898995 still not sure if this is technically accurate

Homework Statement Evaluate: \int\\{1}/{\sqrt{x^2-1}} dx between -3, -2 I know I'm supposed to use hyperbolic substitution in the question. Homework Equations edit: cosh^2(t) - sinh^2(t) = 1 The Attempt at a Solution let x = -cosht, inside the integral let dx = sinh(t) dt int (...