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stuck on u-substituion

find_the_fun

Active member
Feb 1, 2012
166
I was practicing this question which is
\(\displaystyle \int \frac{cos(5x)}{e^{sin(5x)}} dx\)
let u=sin(5x) then \(\displaystyle \frac{du}{dx}=5cos(5x)\) which can be rewritten as \(\displaystyle \frac{1}{5}du = cos(5x) dx\)
Substituting u in gives \(\displaystyle \frac{1}{5} \int \frac{1}{e^u} du\)

This is where I get messed up. Can't you rewrite \(\displaystyle \frac{1}{e^u}\) as \(\displaystyle e^{-u}\)? and then use ∫ax dx = ax/ln(a) + C to get an answer? Is there really no elementary way to integrate \(\displaystyle e^{-u}\)?
 
Last edited:

DreamWeaver

Well-known member
Sep 16, 2013
337
Re: stuck on u substituion

You're on the right track there, since, by definition

\(\displaystyle \frac{1}{z^a}=z^{-a}\)


Bearing that in mind, how does your solution pan out...?



All the best!

Gethin
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: stuck on u substituion

Yes, you are correct when you state that the given substitution tranforms the indefinite integral into:

\(\displaystyle \frac{1}{5}\int e^{-u}\,du\)

Given that:

\(\displaystyle \frac{d}{du}\left(-e^{-u} \right)=e^{-u}\)

we may then write the integral as:

\(\displaystyle \frac{1}{5}\int\,d\left(-e^{-u} \right)\)

Can you proceed?
 

find_the_fun

Active member
Feb 1, 2012
166
Re: stuck on u substituion

So I need to do u substitution again and use another variable?
Let \(\displaystyle v=-u\) then \(\displaystyle \frac{dv}{du}=-1\) rewriting gives \(\displaystyle -dv=du\)
So \(\displaystyle (-1)\frac{1}{5} \int e^v dv= \frac{-e^v}{5}+C=\frac{-e^{-u}}{5}+C=\frac{-e^{-sin(5x)}}{5}+C\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: stuck on u substituion

Using the form I wrote, no further substitution is needed to integrate, since:

\(\displaystyle \int\,du=u+C\)

Once you apply this, then back-substitute for $u$.