Integration (easier than it looks) (1 Viewer)

[KeypadSDM]

I made these up myself and 2/3 members of my class had no idea...
I just want to see what this forum has to say:

/
|(Tan[x])^n dn
/

/
|(Log[x, e])^-1 dx
/

The second on reads: Log of e to the base x.

 

[spice girl]

For the first one, assuming that x is a constant (cos otherwise we don't have anything about it), then
let u = tanx

I(u^n)dn = u^n/lnu + C = (tanx)^n / ln(tanx) + C

For the second one:

I((log[x,e])^-1)dx = I(lnx/lne)dx ..........change of base

= I(lnx)dx
and integrate this by parts

 

[KeypadSDM]

See, easier than it looks

 

[Mathematician]

huh????????/

i get the second one , but the first one...

Where does log come in?
I(u^n)dn = u^n/(n+1) + C

tanx is a constant

 

[KeypadSDM]

You need the Log[tan[x]] ... watch:

/
|a^xdx
/
=
/
|e^Log[a^x]dx
/
=
/
|e^(xLog[a])dx
/
= e^(xLog[a]) / Log[a] + c
= a^x / Log[a] + c

All logs are base e.

 

[Mathematician]

...

I still dont get it .... :mad1:

what is a ? log{tan[x]}= a ?

And where is log{tan[x]} coming from?

 

[ND]

a is simply a constant. KeypadSDM was just showing how to get a constant to the power of the variable into integratable form.

 

[KeypadSDM]

a = Tan[x]
Log[a] = Log[Tan[x]]

.:

/
|(Tan[x])^ndn
/
= (Tan[x])^n/Log[Tan[x]]

It's identical to integration of 2^x (which your 4 unit teacher should have shown you), just replace 2 with Tan[x]