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How to go about evaluating the following integral?

$$ I = \int \sqrt{ \dfrac {\sin(x-\alpha)} {\sin(x+\alpha)} }\,\operatorname d\!x$$

What I have done so far:

$$ I = \int \sqrt{ 1-\tan\alpha\cdot\cot x }\,\operatorname d\!x$$ Let $ t^2 = 1-\tan\alpha\cdot\cot x $

$$ \begin{align} 2t\,\operatorname d\!t &= \tan\alpha \cdot \csc^2x\,\operatorname d\!x \\ & = \tan\alpha \cdot \Bigg(1 + \Big(\dfrac{1-t^2}{\tan\alpha}\Big)^2\Bigg)\,\operatorname d\!x \\ & = \dfrac{\Big(\tan^2 \alpha + (1-t^2)^2\Big)}{\tan \alpha}dx \end{align}$$

So, from that: $$\begin{align} I &= \int \sqrt{ 1-\tan\alpha\cdot\cot x }\,dx \\ & = \int \dfrac{2t^2\tan\alpha}{\Big(\tan^2 \alpha + (1-t^2)^2\Big)}\, \operatorname d\!t \\ \end{align}$$

What to do next?

Edit: I had thought of doing a substitution: $u = 1-t^2$ but that doesn't work as you need one more $t$ term in the numerator.

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  • $\begingroup$ I guess making $x = 1-t^2$ substitution will make it simpler and in a similar form to $\frac{d tan^{-1}(t)}{dt}$. After that I think the best way to get rid of the powers of $t$ in the numerator would be to do it by part. I didn't so myself but you could have a look in that direction $\endgroup$ – user88595 Aug 13 '13 at 13:49
  • $\begingroup$ That's the problem. You can't make that substitution successfully. If you do so, you need one more $t$ term in the numerator. $\endgroup$ – Parth Thakkar Aug 13 '13 at 13:53
  • $\begingroup$ I am not getting anything with this. Some more hint please! $\endgroup$ – Parth Thakkar Aug 13 '13 at 14:14
  • $\begingroup$ Try using partial fraction but you will have to deal with complex numbers. Write $\tan^2(\alpha) + (1-t^2)^2 = (1-t^2 + i\tan (\alpha))(1-t^2 - i\tan (\alpha))$ $\endgroup$ – user88595 Aug 13 '13 at 14:20
  • $\begingroup$ I don't know integration of complex numbers, really! I think this problem can be solved without using them - it is one of the problems in our problem sheet (and we haven't been taught complex integration yet) $\endgroup$ – Parth Thakkar Aug 13 '13 at 14:22
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Given that,

$$I = \int \sqrt{ \dfrac {\sin(x-\alpha)} {\sin(x+\alpha)} }\,dx$$

multiplying and dividing by $\sqrt{\sin(x-\alpha)}$.

we get,

$$I = \int { \dfrac {\sin(x-\alpha)} {\sqrt{\sin(x+\alpha)\cdot \sin (x-\alpha) }}}\,dx$$

$$I=\int \dfrac{ \sin x\cdot \cos{\alpha }}{\sqrt{\sin^2x-\sin^2\alpha}}\ dx-\int\dfrac{\cos x\cdot \sin\alpha} {\sqrt{\sin^2x-\sin^2\alpha}}\ dx$$(how?)

$$I= \cos{\alpha}\int\dfrac{ \sin x dx}{\sqrt{\sin^2x-\sin^2\alpha}}dx-\sin\alpha\int\dfrac{\cos x dx}{\sqrt{\sin^2x-\sin^2\alpha}}dx$$

For the first integral,make the substituion $\cos x=u$. For the second integral make the substituion $\sin x=v$.

You can take it from here.

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  • 3
    $\begingroup$ Superb! The most elegant solution I've come across so far (without using complex numbers!) :D $\endgroup$ – Parth Thakkar Aug 14 '13 at 12:43
  • $\begingroup$ My pleasure.. :D $\endgroup$ – dajoker Aug 14 '13 at 12:46
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Substitute

$$u = \frac{\sin{(x-\alpha)}}{\sin{(x+\alpha)}}$$

Then, with some algebraic manipulation, we find that

$$dx = \frac{2 du}{\sec^2{\alpha} u^2 + 2 (\tan^2{\alpha}-1) u + \sec^2{\alpha}}$$

so that the integral becomes

$$2 \int du \frac{\sqrt{u}}{\sec^2{\alpha} \, u^2 + 2 (\tan^2{\alpha}-1) u + \sec^2{\alpha}}$$

As for the latter integral, break up into its factors $u-u_{\pm}$, where

$$u_{\pm} = \cos{2 \alpha} \pm i \cos{\alpha}$$

and do a partial fractions decomposition, so the integral becomes

$$\frac{1}{i 2 \cos{\alpha}} \left [ \int du \frac{\sqrt{u}}{u-u_+} - \int du \frac{\sqrt{u}}{u-u_-}\right ]$$

To evaluate each of these integrals, let $u=v^2$ so that

$$\int du \frac{\sqrt{u}}{u-u_+} = 2 \int dv \frac{v^2}{v^2-u_+} = 2 v + 2 u_+ \int \frac{dv}{v^2-u_+}$$

the latter integral taking the form of an inverse hyperbolic tangent. The result I get is

$$\int dx \sqrt{\frac{\sin{(x-\alpha)}}{\sin{(x-\alpha)}}} = \frac{1}{\cos{\alpha}} \Im{\left [\sqrt{u_+} \log{\left ( \frac{\sqrt{u} - \sqrt{u_+}}{\sqrt{u} + \sqrt{u_+}}\right)}\right]} + C$$

where, again

$$u = \frac{\sin{(x-\alpha)}}{\sin{(x-\alpha)}} $$ $$u_{+} = \cos{2 \alpha} + i \cos{\alpha}$$

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  • $\begingroup$ As I said, I have not clue about complex integration. Could you please provide an alternative? $\endgroup$ – Parth Thakkar Aug 13 '13 at 15:28
  • $\begingroup$ @ParthThakkar: This is not complex integration. This is simply a manipulation that involves complex numbers. Believe me, the avoidance of such will be way, way more complicated. The integration is over a real variable; the imaginary factor is just that, a factor like any other. $\endgroup$ – Ron Gordon Aug 13 '13 at 15:29
  • $\begingroup$ My bad, didn't read properly $\endgroup$ – Parth Thakkar Aug 13 '13 at 15:38
  • $\begingroup$ What's that J-like thing? (Edit: Is it $Im$?) $\endgroup$ – Soham Chowdhury Aug 13 '13 at 15:40
  • $\begingroup$ @SohamChowdhury: sorry, that is the imaginary part. In LaTeX, it is written as \Im{}. Yeah, that's not the most obvious thing, is it? $\endgroup$ – Ron Gordon Aug 13 '13 at 15:40
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} I&\equiv\int\root{\sin\pars{x - \alpha} \over \sin\pars{x + \alpha}}\,\dd x =\int\root{\sin\pars{x}\cos\alpha - \cos\pars{x}\sin\pars{\alpha}\over \sin\pars{x}\cos\alpha + \cos\pars{x}\sin\pars{\alpha}}\,\dd x \\[3mm]&=\int\root{\tan\pars{x} - \beta\over \tan\pars{x} + \beta}\,\dd x\quad\mbox{where}\quad\beta \equiv\tan\pars{\alpha} \end{align}

\begin{align} I&=\ \overbrace{\int\root{\tan\pars{x} - \beta\over \tan\pars{x} + \beta}\,\dd x} ^{\ds{\mbox{Set}\ x \equiv t/2\ \imp\ t = 2x}}\ =\ \half\ \overbrace{\int\root{\tan\pars{t/2} - \beta\over \tan\pars{t/2} + \beta}\,\dd t} ^{\ds{\mbox{Set}\ y \equiv \tan\pars{t/2}\ \imp\ t = 2\arctan\pars{y}}} \\[3mm]&=\half\int\root{y - \beta \over y + \beta}\,{2\,\dd y \over 1 + y^{2}} \end{align}

With $\ds{{y - \beta \over y + \beta} \equiv z}$: \begin{align} I&=2\beta\int {z\,\dd z \over \pars{\beta^{2} + 1}z^{2} + 2\pars{\beta^{2} - 1}z + \beta^{2} + 1} \\[3mm]&={2\beta \over \beta^{2} + 1}\int {z\,\dd z \over z^{2} + 2\bracks{\pars{\beta^{2} - 1}/\pars{\beta^{2} + 1}}z + 1} =\sin\pars{2\alpha} \int {z\,\dd z \over z^{2} - 2\cos\pars{2\alpha}z + 1} \end{align} You can take it from here.

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