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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.

  • 6
    $\cos x\over \sin x$ is $\cot x$, not $\tan x$.2012-06-03
  • 1
    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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$$\begin{align} \int \frac{\cos x}{\sin x}\mathrm dx +2 \int \frac{\sin x\cos x}{\sin x}\mathrm dx &= \int \cot x\,\mathrm dx+2\int \cos x\,\mathrm dx\\ &= \ln|\sin x|+2\sin x+c \end{align}$$

  • 1
    Hello, welcome to Math.SE. Thank you for your answer! I've edited it to use MathJax. For 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