# Evaluate $\int \frac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$

I`m trying to solve this integral and I did the following steps to solve it but don't know how to continue. $$\int \dfrac{1+\cos(x)}{\sin^2(x)}\,\operatorname d\!x$$
\begin{align}\int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}\,\operatorname d\!x &= \int \dfrac{\operatorname d\!x}{\sin^2(x)}+\int \frac{\cos(x)}{1-\cos^2(x)} \\ &=\int \sin^{-2}(x)\,\operatorname d\!x + \int \cos(x)\,\operatorname d\!x - \int \frac{\operatorname d\!x}{\cos(x)}\end{align} Any suggestions how to continue?
Thanks!

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Avoid writing $\dfrac{1}{\sin^2 x}$ as $\sin ^{-2} x$. – Inceptio May 16 '13 at 10:16
err... $\frac1{a-b} \ne \frac1a - \frac1b$. – Lord_Farin May 16 '13 at 10:17
Ah, didnt see that, realy bad. – Ofir Attia May 16 '13 at 10:20

$$\int \frac{1+\cos(x)}{\sin^2(x)}dx=\int \frac{dx}{\sin^2(x)}+\int \frac{\cos(x)}{\sin^2(x)}dx$$

$$=\int \csc^2xdx+\int\csc x\cot xdx=-\cot x-\csc x+C$$

Alternatively, $$\int \frac{1+\cos(x)}{\sin^2(x)}dx=\int \frac{1+\cos(x)}{1-\cos^2(x)}dx=\int \frac{dx}{1-\cos x}$$

$$\text{Use }\cos x=\frac {1-\tan^2\frac x2}{1+\tan^2\frac x2}$$ and put $\tan\frac x2=u$ (Weierstrass substitution formulas)

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do you have maybe another way ? because I dont use $csc$. thanks – Ofir Attia May 16 '13 at 10:28
@OfirAttia, find the alternative method – lab bhattacharjee May 16 '13 at 10:34
thanks I see that, you use there in universal integration? – Ofir Attia May 16 '13 at 10:35
@OfirAttia, what is universal integration? – lab bhattacharjee May 16 '13 at 10:38
Maybe that how its called here or I dont know, but as you say if $tan(\frac{x}{2}) = u$ so $$cos(x)=\frac{1-t^2}{1+t^2}$$ $$sin(x)=\frac{2t}{1+t^2}$$ $$\tan(x)=\frac{2t}{1-t^2}$$ $$x=2arctan(t)$$ $$dx=\frac{2}{1+t^2}dt$$ – Ofir Attia May 16 '13 at 10:42

Hints:

$1+\cos x=2 \cos^2 \dfrac{x}{2}$

$\sin^2 x=(2\sin\dfrac{x}{2} \cos \dfrac{x}{2})^2$

You expression will be $\dfrac{1}{2} \int \dfrac{1}{\sin^2\dfrac{x}{2}}$

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Hints:

$$(1)\;\;\int\frac{1+\cos x}{\sin^2x}dx=\int\frac1{\sin^2x}dx+\int\frac{\cos x}{\sin^2x}dx$$

$$(2)\;\;\;\;\;\;(\cot x)'=-\frac1{\sin^2x}$$

$$(3)\;\;\;\text{for a derivable function}\;f : \int\frac{f'(x)}{f(x)^m}dx=\frac{f(x)^{1-m}}{1-m}+C\;,\;\;\color{blue}{m\neq -1}$$

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