\(\displaystyle \int\frac{\cos(x)}{e^{2\sin(x)}+4e^{\sin(x)}+4}\ dx\)

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\(\displaystyle \int\frac{\cos(x)}{e^{2\sin(x)}+4e^{\sin(x)}+4}\ dx\)

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Have a look at this That link should help you solve this if you know about it,

\(\displaystyle \int\frac{\cos(x)}{e^{2\sin(x)}+4e^{\sin(x)}+4}\ dx\)

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:

\(\displaystyle \tag(1)u=e^{\sin(x)}\ \ \ \ \frac{du}{dx}=\cos(x)e^{\sin(x)}\ \ \ \ \frac{du}{e^{\sin(x)}}=\cos(x)\ dx\)

\(\displaystyle \tag(2) \int\frac{1}{u(u^2+4u+4)}\ du\)

\(\displaystyle \tag(3) \int\frac{1}{u(u+2)^2}\ du\)

\(\displaystyle \tag(4)\frac{A}{u}+\frac{B}{u+2}+\frac{C}{(u+2)^2}\)

\(\displaystyle \tag(5)\frac{A(u+2)^2+B(u)(u+2)+C(u)}{(u)(u+2)^2}\)

\(\displaystyle \tag(6)A=\frac14\ \ \ \ B=\frac{-1}{4}\ \ \ \ C=\frac{-1}2\)

\(\displaystyle \tag(7)\int\frac{\frac14}{u}+\frac{\frac{-1}4}{u+2}+\frac{\frac{-1}2}{(u+2)^2}\ du\)

\(\displaystyle \tag(8)\frac{\ln|u|}{4}-\frac{\ln|u+2|}{4}+\frac{1}{2u+4}+C\)

\(\displaystyle \tag(9)\frac{\ln|e^{\sin(x)}|}{4}-\frac{\ln|e^{\sin(x)}+2|}{4}+\frac{1}{2e^{\sin(x)}+4}+C\)

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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.)

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Using replies #2 & 3 with the link in #3, see if there is not a rather simple approach.