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I am trying to find the integral of $\int \frac{\cos x+\sin 2x}{\sin x}$

$\int \frac{\cos x}{\sin x} + \int \frac{\sin 2x}{\sin x}$

$\int \tan x + \int \frac{\sin 2x}{\sin x}$

I think I am suppose to have the integral of tanx memorized so I will put that to the side for now.

$\int \frac{\sin 2x}{\sin x}$

I do not know what to do with this since I can't make a u subsitution or anything else so I will just randomly use the double angle identity I have memorized.

$\int \frac{2\sin x\cos x}{\sin x}$

$\int 2\cos x$

$2 \int \cos x$

$2\sin x + \int \tan x$

$2\sin x + \ln|\sec x| + c$

This is of course wrong.

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    I think you are missing a few $dx$'s2013-01-03

3 Answers 3

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Remember that $\dfrac{\cos(x)}{\sin(x)} = \cot(x)$ and not $\tan(x)$. An easier way to do $\int \dfrac{\cos(x)}{\sin(x)} dx$ is to do as follows. Hence, $I = \int \dfrac{\cos(x)}{\sin(x)} dx.$ Set $\sin(x) = t$, then we get $\cos(x) dx = dt$. Hence, $I = \int\dfrac{dt}{t} = \log(t) + C = \log(\lvert \sin(x) \rvert) + C$ Hence, your answer is $2 \sin(x) + \log(\lvert \sin(x) \rvert) + C$

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Hint

Recall the chain rule $g(f(x))'=g'(f(x))\cdot f'(x)$ and note that $\frac{\cos x }{\sin x} =\frac{f'(x) }{f(x)} =\frac{f'(x) }{f(x)}=\frac{1 }{f(x)}\cdot f'(x)$

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$\begin{align} \int \frac{\cos x}{\sin x}\mathrm dx +2 \int \frac{\sin x\cos x}{\sin x}\mathrm dx &= \int \cot x\,\mathrm dx+2\int \cos x\,\mathrm dx\\ &= \ln|\sin x|+2\sin x+c \end{align}$

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    Hello, welcome to Math.SE. Thank you for your answer! I've edited it to use MathJax. $F$or some basic information about writing maths at this site see e.g. [here](http://meta.math.stackexchange.com/questions/5020/), [here](http://meta.stackexchange.com/a/70559/155238), [here](http://meta.math.stackexchange.com/questions/1773/) and [here](http://math.stackexchange.com/editing-help#latexhelp/notation).2013-10-17