Vi Nguyen
May 7, 2009, 07:52 AM
Can someone tell me how to evaluate the integral:
∫tan(2x)sec³(2x)dx
I don't know what to do when sec is cubed and which substitution to make, or if there even is a substitution, if I express sec³(2x) as sec²(2x)sec(2x) and change sec²(2x) to (tan²(2x)+1), then let u=tan(2x) du/dx=sec²(2x) but I only have a sec(2x) left to cancel, :(
∫tan(2x)sec³(2x)dx
I don't know what to do when sec is cubed and which substitution to make, or if there even is a substitution, if I express sec³(2x) as sec²(2x)sec(2x) and change sec²(2x) to (tan²(2x)+1), then let u=tan(2x) du/dx=sec²(2x) but I only have a sec(2x) left to cancel, :(