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#### 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 ∫a

\(\displaystyle \int \frac{cos(5x)}{e^{sin(5x)}} dx\)

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 ∫a

^{x}dx = a^{x}/ln(a) + C to get an answer? Is there really no elementary way to integrate \(\displaystyle e^{-u}\)?
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