Wtf? (1 Viewer)

[aero135]

How do you do this question?

Integrate:

limits are ln 2 and 0

e^x ,
4 + e^2x

using u = e^x

 

[azureus88]

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]

Dont forget to change limits to 1 and 2...

 

[azureus88]

woops, thought it was an indefinite integral.

 

[aero135]

is that inverse functions?

cause we havent done it!

 

[Timothy.Siu]

is that inverse functions?

cause we havent done it!

well u dont need inverse functions, u cud make another substitution after that

 

[azureus88]

[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]

You can just use the table of standard integrals without knowing anything about inverse functions...