Posted by Ryan on August 29, 2002 at 14:50:22:
In Reply to: Need help desperately - seems like a simple integral? posted by Dartchen on August 29, 2002 at 12:00:30:
ill give your prob. a try--> I have been teaching myself calculus so I am not sure if my answer is right....use your own judgement!
integral ( tan(x^1/2))
u= x^1/2
du = (1/2)x^(-1/2) dx
2du = x^-1/2 dx
2u^2 du = dx
2* integral [( u^2)*(tan u ) du ]
2 * (1/3)u^3 tan u + u^2 ln |sec u | + C