<p>Ugh, I know this is probably easy and I'm making some kind of error because I'm not considering the best of methods...but here it is. </p>
<p>Integrate: 1/(csc x - 1) dx
I multiplied by the conjugate (cscx+1) to both numerator and denominator, and then I substituted the denominator with a trig identiy cot^2x, and then seperated the two values in the numerator. I don't know where to go from there.</p>
<p>Uh, I think that's too complicated. Use u-substitution for x-1 and then you get the integral of 1/csc u du, which is ln (csc u). You replace u with x-1 and get ln (csc (x-1)). I'm not sure about this though.</p>
<p>It's not csc(x-1), it's csc(x) - 1. You can't do that.</p>
<p>Even if it was, the integral of 1/csc u du is not ln (csc u).</p>
<p>This doesn't seem too easy. Unless I am forgetting some major substitution thing, I can't see how you do this without IBP.</p>
<p>How would you do it by IBP?</p>
<p>You were going about it the right way. Once you split it up and get the integrals of csc(x)/(cotx)^2 and 1/(cotx)^2, you can change csc(x)/(cotx)^2 to 1/sinx * 1/((cos(x))^2)/(sin(x))^2) which simplifies to sinx/(cosx)^2 or the integral of secxtanx which is just secx. The other part becomes the integral of 1/(cotx)^2 or the integral of sin^2/cos^2 and then I got lost...sorry. I think there's a simple way to go about it but I have AP Chem to study for! lol.</p>
<p>u started off the right way</p>
<p>(csc x + 1) / cot^2x</p>
<p>= (1/sinx + 1)/cot^2x</p>
<p>= (1/ sinx) / cot^2x + 1/cot^2x</p>
<p>= (1/sinx)*(sin^2x/cos^2x) + tan^2x</p>
<p>= sin x / cos^2x + tan^2x</p>
<p>the first part can be intergrated easily (- 1/cos x )</p>
<p>2nd part</p>
<p>tan^2x = tan^2x+1-1 =sec^2x -1</p>
<p>sec^2x's the integral of tan x, -1's just -x</p>
<p>final answer: - 1/cos x + tan x -x</p>
<p>hope i did it correctly. just finished calc review packet</p>
<p>btw i don't think this prob.'s easy by any means; if it is considered easy i'd hate to see the hard ones. :(</p>
<p>I think the x is right, not sure about the second part though.</p>
<p>sec^2x's the integral of tan x</p>
<p>eh i meant the derivative -_-</p>
<p>what's IBP btw? integral by p______?</p>
<p>Integral by parts.</p>
<p>CRAP!!! i dont remember any of the trig identities.....i have calc in 2 weeks.....sob...</p>
<p>LOL you'll learn. Not remembering Trig identities should be the least of the problem.</p>
<p>I'll learn, yes, but not through my teacher. I have the same teacher as I had for pre-calc and he=nutcase. he least he grades easily</p>
<p>is that calculus ab or calculus bc? b/c we just did that type of integration last sememster in bc, it looks like an impromper integral to me. maybe i'm wrong, but i'm just curious, bc we didn't touch integrals anywhere near that difficult in ab.</p>