Hey infraRed.
One that comes to mind hint-wise is to multiply the inside by (1+sin(t))/(1+sin(t)) which will make the denominator 1 - sin^2(t) = cos^2(t) and this will simplify to sec^2(t) - tan^2(t) which is integrable using standard results.
Hi. I'm working through some review questions in prep for the mid-term. This one has me scratching my head a little bit:
I would appreciate any hints. I'm not sure how to apply here the normal approach of using identies and substitution. I played around with a few techniques like adding one, and IBP, but this didn't seem to help.
Hey infraRed.
One that comes to mind hint-wise is to multiply the inside by (1+sin(t))/(1+sin(t)) which will make the denominator 1 - sin^2(t) = cos^2(t) and this will simplify to sec^2(t) - tan^2(t) which is integrable using standard results.
An alternative method for dealing with integrals involving trig functions like this is to use the half-angle substitution
That gets you and
For this particular example it's not as neat as the routine already demonstrated, but there are times when it's the thing to do.