# stuck on u-substituion

#### find_the_fun

##### Active member
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 $$\displaystyle e^{-u}$$?

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#### DreamWeaver

##### Well-known member
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

Staff member
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
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

$$\displaystyle \int\,du=u+C$$
Once you apply this, then back-substitute for $u$.