Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
1  
Avoid writing $\dfrac{1}{\sin^2 x}$ as $\sin ^{-2} x$. –  user63477 May 16 '13 at 10:16
1  
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
add comment

3 Answers

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

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


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)

share|improve this answer
    
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
show 5 more comments

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 $\dfrac{1}{2} \int \dfrac{1}{\sin^2\dfrac{x}{2}}$

share|improve this answer
add comment

Hints:

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

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

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.