Evaluate $\int \sqrt{ \frac {\sin(x-\alpha)} {\sin(x+\alpha)} }\,\operatorname d\!x$?

Solution 1:

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.

Solution 2:

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

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