yfw? Edit... Never mind. Some basic integration is slowly coming back from the dim dark depths of my memory.
Following this through leads me to: \(\int \frac{1}{\cos x} dx = \ln \frac{1 + \sin x}{\1 - \sin x} + C\) Is that right?
Looking at post #2... If we put \(t=\tan (x/2)\) then \(\sin(x)=\frac{2t}{1+t^2}, \cos(x) = \frac{1-t^2}{1+t^2}\) and \(dx = \frac{2}{1+t^2} dt\) so \(\int \frac{1}{\cos x}dx = \int \frac{1+t^2}{1-t^2}\frac{2}{1+t^2}dt\) \(=\int \frac{2}{1-t^2}dt\) Not sure where to go from here...
Ah! Thanks temur. So: \(\int \frac{2}{1-t^2}dt = \int \frac{1}{1-t} + \frac{1}{1+t}~dt\) \(=-\ln|1-t| + \ln|1+t| +c = \ln\left|\frac{1+t}{1-t}\right|+c\) \(=\ln\left|\frac{1+\tan(x/2)}{1-\tan(x/2)}\right| + c\) Now, can we express this as in post #2?
For the numerator, \(\frac1{\sqrt2}\cos(x/2)+\frac1{\sqrt2}\sin(x/2)=\sin(\pi/4)\cos(x/2)+\cos(\pi/4)\sin(x/2)=\sin(x/2+\pi/4)\).
Well, I simply prefer my answer. I always liked cos over sin. Sin is a jerk, and cos is cool. And I try never to use tan, because it causes skin cancer.
Hey guys thanks for your posts Please Register or Log in to view the hidden image! So how many different answers did we come up with?
I'm the same way. My mathematics degree doesn't mean I can remember integral tables. I know the theory behind it all, but my memory is terrible. People laugh at me when I fail to do basic math... It doesn't bother me though. Besides, I had a professor that was brilliant when it came to numerical linear algebra, but oddly enough she has dyscalculia.
\(\int \sec x dx = \int \sec x {(\sec x + \tan x) \over (\sec x +\tan x)}dx= \int {\sec^2 x + \sec x \tan x \over \sec x +\tan x}dx\) u = sec x + tan x
