I have two problems that I'm having trouble with. Both of them involve square roots in the denominator. I'm not sure how to deal with that. Also, I'm new to the forum so I'm not sure how to make all the math symbols appear. Please bear with me!
1) integrate (1/(x^2-r^2)^(1/2))dx assuming r is some constant
2) integrate ((x+2)/((x^2+4x+5)^(1/2)))dx from 0 to infinity
I'm really stuck, so I'd appreciate any help!!
The u-substitution for #2 worked very well. I obtained sqrt(x^2+4x+5)+c, as an indefinite solution. From 0 to infinity, I got, infinity-sqrt(5). (which I figure is still infinity, but I guess I'll write in the sqrt(5) for completeness)
I'm still confused about #1 though. This is what I have so far. Using the trig substitution, x=r*sec(theta); dx=rsec(theta)tan(theta)d(theta). However, I'm unsure what to substitute 1/sqrt(x^2-r^2) with. I thought maybe (sin(theta))/x. But I don't know how that helps?
That was a ton of help! Thank you!! I still ran into a rough spot...and I'll show you where it got messy, so I'm not completely sure that my answer is 100% right.
so, starting from where o_O left off, the integral of r*sec(theta)*tan(theta)d(theta)/sqrt((r^2)(sec^2(theta)-1) = the integral of (r*sec(theta)*tan(theta)d(theta))/sqrt((r^2)-tan^2(theta) = the integral of (r*sec(theta)*tan(theta)d(theta))/(r*tan(theta)) = the integral of sec(theta)d(theta)
at which point I googled some help, which gave me a very detailed solution, that I probably wouldn't have thought of on my own.
getting back to my eventual answer, I end up with ln|sec(theta) + tan(theta)| + C. But then I need to convert back to x and r.
I find that theta = arcsec(x/r), through the first trig substitution that Chris L T521 gave me.
substituting back in, gives me ln|(x/r) + tan(arcsec(x/r)|+C
I'm not sure that this is the simpliest form?
Whew that was long..I really need to figure out how to work these math symbols so it doesn't look so messy....Thanks again!
This means that
Now when we're dealing with right angle trigonometry, this means that the hypotenuse will have a length of and the side adjacent to the angle will have a length of . Thus, the side opposite the angle will have a length of
So the tangent of this angle will be opposite over adjacent, or
So you could write your answer as
Does this make sense?