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In a related question the following integral was evaluated $$ \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{\mathrm{d}x/2}{1 + \cos x \sin x} =\int_0^{2\pi} \frac{\mathrm{d}x/2}{2 + \sin x} \,\mathrm{d}x =\int_{-\infty}^\infty \frac{\mathrm{d}x/2}{1+x+x^2} $$ I noticed something interesting, namely that $$ \begin{align*} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x & = \int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\ & = \int_0^{\pi} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x = \int_0^{\pi} \frac{(\sin x)^2}{1 - \cos x \sin x} \,\mathrm{d}x \end{align*} $$ The same trivially holds if the upper limits are changed to $\pi/2$ as well ($x \mapsto \pi/2 -u$). But I had problems proving the first equality. Does anyone have some quick hints?

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    $\begingroup$ consider evaluating $$\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} - \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x = \int_0^{\pi} \frac{\cos (2x)}{1 +\frac{1}{2}\sin(2 x)}\,\mathrm{d}x $$ $\endgroup$ – ganeshie8 Oct 2 '14 at 10:27
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    $\begingroup$ Are you not satisfied yet with the answers below? $\endgroup$ – Anastasiya-Romanova 秀 Oct 7 '14 at 12:07
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    $\begingroup$ @N3buchadnezzar Oh sorry, I didn't know that. Get well soon ᕙ( ^ₒ^ c) $\endgroup$ – Anastasiya-Romanova 秀 Oct 8 '14 at 8:21
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    $\begingroup$ Someday ;)${}{}{}$ $\endgroup$ – N3buchadnezzar Oct 8 '14 at 10:12
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    $\begingroup$ math.stackexchange.com/questions/61605/… If you want further discussion ping me in chat. $\endgroup$ – N3buchadnezzar Oct 8 '14 at 10:32
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Split the integral into two terms with limit $\left[0,\frac{\pi}{2}\right]$ and $\left[\frac{\pi}{2},\pi\right]$ \begin{align} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x +\int_{\pi/2}^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \\ \end{align} Using identity \begin{align} \int_a^{b} f(x) \,\mathrm{d}x = \int_a^{b} f(a+b-x) \,\mathrm{d}x \end{align} Also the facts that $\cos\left(\frac{\pi}{2}-x\right)=\sin x$, $\sin\left(\frac{\pi}{2}-x\right)=\cos x$, $\cos\left(\frac{3\pi}{2}-x\right)=-\sin x$, and $\sin\left(\frac{3\pi}{2}-x\right)=-\cos x$, we get \begin{align} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x &=\int_0^{\pi/2} \frac{(\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x +\int_{\pi/2}^{\pi} \frac{(-\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x \\ &=\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\tag{1} \end{align}


Let \begin{align} I=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \end{align} Since \begin{align} \int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x \end{align} then \begin{align} 2I&=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x +\int_0^{\pi} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\\ &=\int_0^{\pi} \frac{(\cos x)^2 + (\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x\\ I&=\frac{1}{2}\int_0^{\pi} \frac{1}{1 + \cos x \sin x} \,\mathrm{d}x\tag{2} \end{align}


Using $(2)$, we get \begin{align} I&=\int_0^{\pi} \frac{1}{2 + 2\cos x \sin x} \,\mathrm{d}x\\ &=\int_0^{\pi} \frac{1}{2 + \sin (2x)} \,\mathrm{d}x\qquad\Rightarrow\qquad x\mapsto2x\\ &=\frac{1}{2}\int_0^{2\pi} \frac{1}{2 + \sin x} \,\mathrm{d}x\tag{3} \end{align}

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Subtract the two integrals in question and get

$$\int_0^{\pi} dx \frac{\cos{2 x}}{1+\frac12 \sin{2 x}} = \frac12 \int_0^{2 \pi} du \frac{\cos{u}}{1+\frac12 \sin{u}}$$

This may be shown to be equal to the complex integral

$$-\frac{i}{4} \oint_{|z|=1} \frac{dz}{z} \frac{z+z^{-1}}{1+\frac{1}{4 i} (z-z^{-1})} = \oint_{|z|=1}\frac{dz}{z} \frac{z^2+1}{z^2+i 4 z-1}$$

The poles of the integrand within the unit circle are at $z=0$ and $z=-(2-\sqrt{3}) i$; their respective residues are $-1$ and $1$. By the residue theorem, therefore, the integral is zero and the two original integrals are equal.

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A quick hint is noticing the symmetry. A rigorous proof is that $$\int\limits_a^b {\frac{{f(\cos x)}}{{g(\sin x\cos x)}}dx} = \int\limits_a^b {\frac{{f\left( {\sin \left( {\frac{\pi }{2} - x} \right)} \right)}}{{g\left( {\cos \left( {\frac{\pi }{2} - x} \right)\sin \left( {\frac{\pi }{2} - x} \right)} \right)}}dx} = \int\limits_{\frac{\pi }{2} - b}^{\frac{\pi }{2} - a} {\frac{{f\left( {\sin u} \right)}}{{g\left( {\cos u\sin u} \right)}}du} $$Hope it helps ;)

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$$ \frac{\cos(\pi/2-x)^2}{1+\cos(\pi/2-x)\sin(\pi/2-x)} = \frac{\sin(x)^2}{1+\sin(x)\cos(x)} $$

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The integrands are both periodic with period $\pi$, so it suffices to verify the identity over any interval of length $\pi$, including $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Then, since both integrands are even functions, it suffices to check that the identity holds when integrating over $\left[0, \frac{\pi}{2}\right]$, that is, that $$ \int_0^{\frac{\pi}{2}} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x =\int_0^{\frac{\pi}{2}} \frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x.$$

But one can show this simply by substituting $x = \frac{\pi}{2} - u$ on either side.

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  • $\begingroup$ Your answer was the first I got, andthe first one I used. However for later readers who also stumble uppon this integral I feel the answer given by Anastasiya is clearer. $\endgroup$ – N3buchadnezzar Oct 7 '14 at 17:30
  • $\begingroup$ The important thing is that you've learned something from posting here, which it seems you have. $\endgroup$ – Travis Oct 8 '14 at 3:23
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$$I=\int_0^{\pi} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$ $$J=\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$


Use $\int_a^b f(x)dx=\int_a^nf(a+b-x)dx$ and $\int_0^{2a}f(x)dx=\int_0^a f(x)dx+f(2a-x)dx$


$$I=\int_0^{\pi/2} \frac{(\cos (\pi-x))^2}{1 + \cos (\pi-x) \sin (\pi-x)} \,\mathrm{d}x+\frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$ $$I=\int_0^{\pi/2} \frac{(\cos x)^2}{1 - \cos x \sin x} \,\mathrm{d}x+\frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$ $$I=\int_0^{\pi/2} \frac{(\cos (\pi/2-x))^2}{1 - \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x+\frac{(\cos (\pi/2-x))^2}{1 + \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x$$ $$I=\int_0^{\pi/2} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x+\frac{(\sin x)^2}{1 + \cos x \sin x} \,\mathrm{d}x$$ $$I=\int_0^{\pi/2} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x+\frac{(\sin (\pi-x))^2}{1 - \cos (\pi-x) \sin(\pi-x)} \,\mathrm{d}x$$ $$I=\int_0^{\pi} \frac{(\sin x)^2}{1 - \sin x \cos x} \,\mathrm{d}x$$


Hope you can now show that: $$J=\int_0^{\pi/2} \frac{(\cos x)^2}{1 + \cos x \sin x} \,\mathrm{d}x=\int_0^{\pi/2} \frac{(\cos (\pi/2-x))^2}{1 + \cos (\pi/2-x) \sin (\pi/2-x)} \,\mathrm{d}x=\int_0^{\pi/2} \frac{(\sin x)^2}{1 + \sin x \cos x} \,\mathrm{d}x$$

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