How do you integrate:
Calculus - bobpursley, Wednesday, January 9, 2013 at 7:56pm
draw a right triangle, x as the hypotensue, 1 as the adjacent side to angle theta, and the opposite side is sqrt (x^2-1)
-sintheta* dT=-1/x^2 dx or
dx= x^2 sinTheta dT
the integral is now...
INt 1/tanTheta*x^2 sinTheta dT
INT cosT/SinT* sec^2T*sinT dT
INT= secT dT=ln(secT+tanT)
check all that, I did it in my head.
Calculus - Kyle, Wednesday, January 9, 2013 at 8:17pm
Hm, I've never done integrals using that method before. I'll check it.