Stuck on U-Substitution: Find Answer

[find_the_fun]

I was practicing this question which is
[math]\int \frac{cos(5x)}{e^{sin(5x)}} dx[/math]
let u=sin(5x) then [math]\frac{du}{dx}=5cos(5x)[/math] which can be rewritten as [math]\frac{1}{5}du = cos(5x) dx[/math]
Substituting u in gives [math]\frac{1}{5} \int \frac{1}{e^u} du[/math]

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

 

[DreamWeaver]

Re: stuck on u substituion

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

$$\frac{1}{z^a}=z^{-a}$$Bearing that in mind, how does your solution pan out...?
All the best!

Gethin

 

[MarkFL]

Re: stuck on u substituion

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

$$\frac{1}{5}\int e^{-u}\,du$$

Given that:

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

we may then write the integral as:

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

Can you proceed?

 

[find_the_fun]

Re: stuck on u substituion

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

 

[MarkFL]

Re: stuck on u substituion

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

$$\int\,du=u+C$$

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