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Solving the Indefinite integral: $$\int\frac{2\cos^2x+2}{\sin^3x}\, \operatorname{d}\!x$$

What's the proper way of finding Indefinite integral in this one? can't get rid of either $\cos$ or $\sin$.

Thanks.

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  • $\begingroup$ Don't forget that, if you are integrating with respect to $x$ then you must use $\operatorname{d}\!x$. $\endgroup$ – Fly by Night Jul 30 '13 at 15:31
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Let $u=\cos x$ so $du=-\sin x dx$ and we have $$\int\frac{2\cos^2 x+2}{\sin^3x}dx=-2\int\frac{u^2+1}{(1-u^2)^2}du=-\int\frac{du}{(1-u)^2}-\int\frac{du}{(u+1)^2}\\=-(1-u)^{-1}+(1+u)^{-1}+C=\frac{1}{1+\cos x}-\frac{1}{1-\cos x}+C=-2\frac{\cos x}{\sin^2 x}+C$$

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  • $\begingroup$ You might continue with combining and simplifying the $\cos$ terms. $\endgroup$ – marty cohen Jul 30 '13 at 15:53
  • $\begingroup$ @martycohen I edited my answer as you suggested but the simplification is a matter of taste and maybe there are people who prefer write the final result with the function $\mathrm{cotan}$ :) $\endgroup$ – user63181 Jul 30 '13 at 17:39
  • $\begingroup$ @SamiBenRomdhane: See the general case below. ;-) $\endgroup$ – mrs Aug 1 '13 at 9:00
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A good point for the next. Assume you have the following integral: $$ \int R(\sin x,\cos x)dx $$

If you have $$R(-\sin x,\cos x)\equiv-R(\sin x,\cos x)$$ then it is easier for us to set $\cos x=t$. And you can see this fact is satisfied for the integrand.

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  • $\begingroup$ In fact I used this method that you mentioned and it's called Bioche's rules in the jargon of French mathematics (I have not seen it in English) see fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche and I hope you do not find a problem in the French language:) $\endgroup$ – user63181 Aug 1 '13 at 10:09
  • $\begingroup$ @SamiBenRomdhane: Thanks for remarking that link. I'll try to get along French one. :-) $\endgroup$ – mrs Aug 1 '13 at 10:13
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$$\int \frac{4-2\sin^{2}(x)}{\sin^{3}(x)} \ dx = 4\cdot \int\csc^{3}(x)\ dx -2 \int \csc(x) \ dx $$

Now you can integrate $\csc^{3}(x)$ by writing it as $\csc^{2}(x) \cdot \csc(x) = (\cot^{2}(x)-1)\cdot \csc(x)$

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$$ \begin{aligned} I&=\int\frac{2\cos^2 x + 2}{\sin^3 x}\,\mathrm{d}x\\&=2\int\frac{1+\cos^2 x}{\sin^4 x}\sin x\,\mathrm{d}x\\ &=2\int\frac{1+\cos^2 x}{\left(1 - \cos^2 x\right)^2}\sin x\,\mathrm{d}x\\ &=2\int\frac{1+\cos^2 x}{\left(\cos^2 x-1\right)^2}\sin x\,\mathrm{d}x\\ &=2\int\frac{\cos^2 x\left(1 + \dfrac{1}{\cos^2x}\right)}{\cos^2 x\left(\cos x - \dfrac{1}{\cos x}\right)^{\!2}}\sin x\,\mathrm{d}x\\ &=2\int\frac{1+\dfrac{1}{\cos^2x}}{\left(\cos x - \dfrac{1}{\cos x}\right)^{\!2}}\sin x\,\mathrm{d}x \end{aligned} $$ Now, set $u=\cos x - \dfrac{1}{\cos x}$ and $\mathrm{d}u = -\left(1+\dfrac{1}{\cos ^2 x}\right)\sin x\,\mathrm{d}x$. Then, $$ \begin{aligned} I&=2\int-\dfrac{\mathrm{d}u}{u^2}\\ &=\dfrac{2}{u}+C\\ &=\dfrac{2}{\cos x - \dfrac{1}{\cos x}}+C\\ &=\dfrac{2\cos x}{\cos^2 x - 1} + C\\ &=-\dfrac{2\cos x}{\sin^2 x} + C\\ \end{aligned}\\\\ $$ $$ \boxed{\displaystyle{\int\dfrac{2\cos^2 x + 2}{\sin^3 x}\,\mathrm{d}x = -2\csc x \cot x + C}} $$

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