A aero135 Super Ultimate User! Joined Oct 23, 2007 Messages 242 Gender Male HSC 2008 Mar 7, 2009 #1 How do you do this question? Integrate: limits are ln 2 and 0 e^x , 4 + e^2x using u = e^x
A azureus88 Member Joined Jul 9, 2007 Messages 278 Gender Male HSC 2009 Mar 7, 2009 #2 Substitue u=e^x du=(e^x)dx [maths]\int \frac{du}{4+u^2}=\frac{1}{2}\tan^{-1}\frac{u}{2}[/maths] [maths]=\frac{1}{2}\tan^{-1}\frac{e^x}{2}[/maths] Last edited: Mar 7, 2009
Substitue u=e^x du=(e^x)dx [maths]\int \frac{du}{4+u^2}=\frac{1}{2}\tan^{-1}\frac{u}{2}[/maths] [maths]=\frac{1}{2}\tan^{-1}\frac{e^x}{2}[/maths]
Trebla Administrator Administrator Joined Feb 16, 2005 Messages 8,397 Gender Male HSC 2006 Mar 7, 2009 #3 Dont forget to change limits to 1 and 2...
A azureus88 Member Joined Jul 9, 2007 Messages 278 Gender Male HSC 2009 Mar 7, 2009 #4 woops, thought it was an indefinite integral.
A aero135 Super Ultimate User! Joined Oct 23, 2007 Messages 242 Gender Male HSC 2008 Mar 8, 2009 #5 is that inverse functions? cause we havent done it!
T Timothy.Siu Prophet 9 Joined Aug 6, 2008 Messages 3,447 Location Sydney Gender Male HSC 2009 Mar 8, 2009 #6 aero135 said: is that inverse functions? cause we havent done it! Click to expand... well u dont need inverse functions, u cud make another substitution after that
aero135 said: is that inverse functions? cause we havent done it! Click to expand... well u dont need inverse functions, u cud make another substitution after that
A azureus88 Member Joined Jul 9, 2007 Messages 278 Gender Male HSC 2009 Mar 8, 2009 #7 [maths]\int \frac{du}{4+u^2}[/maths] Substitute u=2tan@ du=(2sec^2@)d@ [maths]=\int \frac{2sec^2\Theta }{4sec^2\Theta }d\Theta [/maths] [maths]=\frac{1}{2}\Theta =\frac{1}{2}\tan^{-1}\frac{u}{2}=\tan^{-1}\frac{e^x}{2}[/maths] That takes ages though, just learn inverse trig.
[maths]\int \frac{du}{4+u^2}[/maths] Substitute u=2tan@ du=(2sec^2@)d@ [maths]=\int \frac{2sec^2\Theta }{4sec^2\Theta }d\Theta [/maths] [maths]=\frac{1}{2}\Theta =\frac{1}{2}\tan^{-1}\frac{u}{2}=\tan^{-1}\frac{e^x}{2}[/maths] That takes ages though, just learn inverse trig.
Trebla Administrator Administrator Joined Feb 16, 2005 Messages 8,397 Gender Male HSC 2006 Mar 8, 2009 #8 You can just use the table of standard integrals without knowing anything about inverse functions...