Integration ∫ [√(sin^2 x-3sin x+2))/√(sin^2 x+3sin x+2))]dx

[juantheron]

Evaluation of $\displaystyle \int \sqrt{\frac{\sin^2 x-3\sin x+2}{\sin^2 x+3\sin x+2}}dx$

 

[juantheron]

Solution [sp]Let $\displaystyle I = \int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$

We can write $\displaystyle \sqrt{\frac{1-\sin x}{1+\sin x}} = \sqrt{\frac{1-\sin x}{1+\sin x}\times \frac{1+\sin x}{1+\sin x}} = \frac{\cos x}{1+\sin x}$So we get $\displaystyle I = \int\frac{\cos x}{1+\sin x}\cdot \sqrt{\frac{2-\sin x}{2+\sin x}}dx$Now Let $1+\sin x= y\;,$ Then $\cos xdx = dy$So Integral $\displaystyle I = \int\frac{1}{y}\cdot \sqrt{\frac{3-y}{1+y}}dy$Now Put $\displaystyle \frac{3-y}{1+y}=t^2\Rightarrow y=\frac{3-t^2}{1+t^2}$So we get $\displaystyle y=-\left[1-\frac{4}{1+t^2}\right] = \left[\frac{4}{1+t^2}-1\right].$

So $\displaystyle dy = -\frac{8t}{(1+t^2)^2}$So Integral $\displaystyle I = \int\frac{1+t^2}{3-t^2}\cdot t\cdot \frac{-8t}{(1+t^2)^2}dt = 8\int\frac{t^2}{(t^2-3)\cdot (1+t^2)}dt$So Integral $\displaystyle I = 2\int \left[\frac{3(t^2+1)+(t^2-3)}{(t^2-3)\cdot (1+t^2)}\right]dt = 2\int \left[\frac{3}{t^2-(\sqrt{3})^2}+\frac{1}{1+t^2}\right]dt$So Integral $\displaystyle I = 6\cdot \frac{1}{2\sqrt{3}}\cdot \ln\left|\frac{t-\sqrt{3}}{t+\sqrt{3}}\right|+2\tan^{-1}(t)+\mathcal{C}$So Integral $\displaystyle I = \sqrt{3}\cdot \ln\left|\frac{\sqrt{2-\sin x}-\sqrt{3}\cdot \sqrt{2+\sin x}}{\sqrt{2-\sin x}-\sqrt{3}\cdot \sqrt{2+\sin x}}\right|+2\tan^{-1}\left(\sqrt{\frac{2-\sin x}{2+\sin x}}\right)+\mathcal{C}$[/sp]