Solving: [MATH]\int\frac{\cos(x)}{e^{2\sin(x)}+4e^{\sin(x)}+4}\ dx[/MATH]

burt

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Aug 1, 2019
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I was given the following problem which I thought I knew how to solve. The only problem is that the answer I got was not given as an option for an answer. It doesn't seem to be equivalent either. It must be that I went wrong somewhere - where though? This is the problem and my steps:
\(\displaystyle \int\frac{\cos(x)}{e^{2\sin(x)}+4e^{\sin(x)}+4}\ dx\)
 

Jomo

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Dec 30, 2014
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burt, how can we find your error if you do not post your work. You could have received help by now if you followed our policy.

In any case, I would use the substitution u = e^sinx and see where that gets you.
 

pka

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Jan 29, 2005
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I was given the following problem which I thought I knew how to solve. The only problem is that the answer I got was not given as an option for an answer. It doesn't seem to be equivalent either. It must be that I went wrong somewhere - where though? This is the problem and my steps:
\(\displaystyle \int\frac{\cos(x)}{e^{2\sin(x)}+4e^{\sin(x)}+4}\ dx\)
Have a look at this That link should help you solve this if you know about it,
 

burt

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Aug 1, 2019
Messages
206
I apologize @Jomo @pka - something happened. This is supposed to be the entire post:
I was given the following problem which I thought I knew how to solve. The only problem is that the answer I got was not given as an option for an answer. It doesn't seem to be equivalent either. It must be that I went wrong somewhere - where though? This is the problem and my steps:
\(\displaystyle \int\frac{\cos(x)}{e^{2\sin(x)}+4e^{\sin(x)}+4}\ dx\)
These are my steps:

\(\displaystyle \tag(1)u=e^{\sin(x)}\ \ \ \ \frac{du}{dx}=\cos(x)e^{\sin(x)}\ \ \ \ \frac{du}{e^{\sin(x)}}=\cos(x)\ dx\)
\(\displaystyle \tag(2) \int\frac{1}{u(u^2+4u+4)}\ du\)
\(\displaystyle \tag(3) \int\frac{1}{u(u+2)^2}\ du\)
\(\displaystyle \tag(4)\frac{A}{u}+\frac{B}{u+2}+\frac{C}{(u+2)^2}\)
\(\displaystyle \tag(5)\frac{A(u+2)^2+B(u)(u+2)+C(u)}{(u)(u+2)^2}\)
\(\displaystyle \tag(6)A=\frac14\ \ \ \ B=\frac{-1}{4}\ \ \ \ C=\frac{-1}2\)
\(\displaystyle \tag(7)\int\frac{\frac14}{u}+\frac{\frac{-1}4}{u+2}+\frac{\frac{-1}2}{(u+2)^2}\ du\)
\(\displaystyle \tag(8)\frac{\ln|u|}{4}-\frac{\ln|u+2|}{4}+\frac{1}{2u+4}+C\)
\(\displaystyle \tag(9)\frac{\ln|e^{\sin(x)}|}{4}-\frac{\ln|e^{\sin(x)}+2|}{4}+\frac{1}{2e^{\sin(x)}+4}+C\)
 

Dr.Peterson

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Nov 12, 2017
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What are the given options? Your work looks good to me.

But look at your first term. Can you simplify it?

(If that were the only issue, then I'm sure you would not say none of the answers are equivalent.)
 

pka

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burt

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Aug 1, 2019
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Ah, I see. My work is correct it was a log rule problem. The first and second terms can be combined.
 
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