Recent content by just_see

Ah... yeah I see. I did that at one point but didn't think to write sec^2 as (1+tan^2). Anyway, thanks a ton.

Hmm.. I just tried a different approach, would like to know if this is a correct approach... \int_{0}^{pi/4}\sec^{2}{x}tan^{6}(x)sec(x)dx - I using u = secx, dv = tan^6xsec^2x du = secxtanx, v = (1/7)tan^7x \frac{sec(x)tan^{7}(x)}{7} - \frac{1}{7}\int_{0}^{pi/4}\sec(x)tan^{8}(x)dx - I =...

Yeah... that gets me: \int_{0}^{pi/4}\sec^{2}{x}tan^{6}(x)sec(x)dx - I u = sec^2xtan^5x dv = secxtanx du = 2sec^2xtan^6x + 5tan^4xsec^4x sec^{3}(x)tan^{5}(x) - \int_{0}^{pi/4}\sec(x)[2sec^{2}(x)tan^{6}(x) + 5tan^{4}(x)sec^{4}(x)]dx simplified... sec^{3}(x)tan^{5}(x) -...

Homework Statement If \int_{0}^{pi/4}\tan^{6}(x)sec(x)dx = I then express \int_{0}^{pi/4}\tan^{8}(x)sec(x)dx in terms of I. Homework Equations The Attempt at a Solution I figured you would have to get the second equation to look like the first one, so I pulled out a tan^2(x) to...

um... cos^2(x)-1 = -sin^2(x)... int[cotxcscxdx] - int[csc^2(x)] -cscx + cotx? Or did I drop something somewhere? (which is quite possible)

Homework Statement \int\frac{dx}{cos(x)-1} Homework Equations None The Attempt at a Solution I've had extreme problems in even attempting to begin this problem. I initially tried to multiply by the conjugate to get int[(cosx+1)/(cos^2(x)-1)], but this doesn't seem to be leading me...