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$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx$

I tried to solve it.

$\int\limits_{0}^{\pi/2}\frac{1+2\cos x}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{4+2\cos x}{(2+\cos x)^2}-\frac{3}{(2+\cos x)^2}dx=\int\limits_{0}^{\pi/2}\frac{2}{2+\cos x}-\frac{3}{(2+\cos x)^2}dx$

But i could not solve further.Please help me in completing.

diya
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3 Answers3

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You may observe that: $$\frac{2 \cos (x)+1}{(\cos (x)+2)^2}=\frac{\cos (x)}{\cos (x)+2}+\frac{\sin ^2(x)}{(\cos (x)+2)^2}=\frac{\frac{d}{dx}\sin(x)}{\cos (x)+2}-\frac{\sin(x)\frac{d}{dx}(\cos x +2)}{(\cos (x)+2)^2}$$

Math-fun
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Given $\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1+2\cos x}{(2+\cos x)^2}dx.\;,$ Now Divide both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $\sin^2 x\;,$ we get

$\displaystyle I = \int_{0}^{\frac{\pi}{2}}\frac{\csc^2 x+2\cot x \cdot \csc x}{\left(2\csc x+\cot x\right)^2}dx\;,$ Now Put $\displaystyle \left(2\csc x+\cot x \right) = t\;,$

Then $\left(\csc^2 x+2\cot x \cdot \csc x \right)dx = -dt.$

So Integral Convert into $\displaystyle I = -\int\frac{1}{t^2}dt = \frac{1}{t} = \left[\frac{\sin x}{2+\cos x }\right]_{0}^{\frac{\pi}{2}} = \frac{1}{2}$

juantheron
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Considering $$I=\int\frac{1+2\cos (x)}{(2+\cos (x))^2}\,dx$$ you can do several things.

First, considering the squared denominator, you could suppose that the result of integration would be something like $$\frac {a+b\sin(x)+c\cos(x)}{2+\cos(x)}$$ which, one differentiated, would give $$\frac{(a-2 c) \sin (x)+2 b \cos (x)+b}{(2+\cos (x))^2}$$ So, $$\frac{(a-2 c) \sin (x)+2 b \cos (x)+b}{(2+\cos (x))^2}=\frac{1+2\cos (x)}{(2+\cos (x))^2}$$ Removing the denominator let us with $$(a-2 c) \sin (x)+2 (b-1) \cos (x)+b-1=0$$ So, $b=1$, $a-2c=0$ makes $$I=\frac{\sin(x)}{2+\cos (x)}$$

The second way would use the tangent half-angle substitution $t=\tan(\frac x2)$ which would lead to $$I=\int\frac{1+2\cos (x)}{(2+\cos (x))^2}\,dx=-\int\frac{2 \left(t^2-3\right)}{\left(t^2+3\right)^2}\,dt$$ which is not very difficult to integrate.

I am sure that you can takle from here.

Claude Leibovici
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  • A plus one for you tonight my friend! I hope all is well. – Mark Viola Aug 13 '15 at 05:42
  • @Dr.MV. Hello ! Here, it is almost 8:00am. Hope you are fine ! – Claude Leibovici Aug 13 '15 at 05:43
  • Bon Jour. Ici, il est minuit quarante cinq. Did I write it correctly? I am OK. For my surgery, the wound is not healing properly and I am undergoing Platelet Rich Plasma treatment in which my peripheral blood is separated in a centrifuge and the rich platelets are injected into the affected site. This should promote growth hormones and stem cells and stimulate healing. I will know in a couple of weeks. – Mark Viola Aug 13 '15 at 05:49
  • @Dr.MV. Your French is perfect. I wish you the best of the best, my friend ! – Claude Leibovici Aug 13 '15 at 05:51
  • Merci! And I wish you the best also. You're one of the really good guys on this site. – Mark Viola Aug 13 '15 at 05:53