M Muteki New member Joined Jun 8, 2007 Messages 8 Jun 18, 2007 #1 dy/sqrt(e^(2y)-1) I know i need to use a u sub, but I cannot seem to find the correct sub for this problem.
dy/sqrt(e^(2y)-1) I know i need to use a u sub, but I cannot seem to find the correct sub for this problem.
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jun 18, 2007 #2 Re: dy/sqrt(e^(2y)-1) Hello, Muteki! \(\displaystyle \L\int\frac{dy}{\sqrt{e^{2y}\,-\,1}}\) Click to expand... Let \(\displaystyle e^y \:=\:\sec\theta\;\;\Rightarrow\;\;y \:=\:\ln(\sec\theta)\;\;\Rightarrow\;\;dy \:=\:\frac{\sec\theta\tan\theta}{\sec\theta}\,d\theta\:=\:\tan\theta\,d\theta\) . . and: \(\displaystyle \:\sqrt{e^{2y}\,-\,1}\:=\:\sqrt{\sec^2\theta\,-\,1} \:=\:\sqrt{\tan^2\theta} \:=\:\tan\theta\) Substitute: \(\displaystyle \L\:\int\frac{\tan\theta\,d\theta}{\tan\theta} \;=\;\int d\theta \;=\;\theta\,+\,C\) Back-substitute: \(\displaystyle \L\:\text{arcsec}\left(e^y\right)\,+\,C\)
Re: dy/sqrt(e^(2y)-1) Hello, Muteki! \(\displaystyle \L\int\frac{dy}{\sqrt{e^{2y}\,-\,1}}\) Click to expand... Let \(\displaystyle e^y \:=\:\sec\theta\;\;\Rightarrow\;\;y \:=\:\ln(\sec\theta)\;\;\Rightarrow\;\;dy \:=\:\frac{\sec\theta\tan\theta}{\sec\theta}\,d\theta\:=\:\tan\theta\,d\theta\) . . and: \(\displaystyle \:\sqrt{e^{2y}\,-\,1}\:=\:\sqrt{\sec^2\theta\,-\,1} \:=\:\sqrt{\tan^2\theta} \:=\:\tan\theta\) Substitute: \(\displaystyle \L\:\int\frac{\tan\theta\,d\theta}{\tan\theta} \;=\;\int d\theta \;=\;\theta\,+\,C\) Back-substitute: \(\displaystyle \L\:\text{arcsec}\left(e^y\right)\,+\,C\)
