I have been comparing these indefinite integrals to the known trig substitution rules and I cannot seem to get them to match. What is the trick?
************************
Well, for example in the second problem, since you have -x^2, your trig substitution would have to be a sin(u), and since you have the 49, you would have to use x=7sin(u). Then dx = 7 cos(u)du and you'd substitute those. The second problem will rearrange tan^2(x) +1 = sec^2(x) to make it sec^2(x)-1=tan^2(x), so that you'd substitute x=7sec(u) and dx = 7sec(u)tan(u)du. Btw, the textbook section for this part of the homework has a thorough treatment of this. I recommend it.
No comments:
Post a Comment