Evaluate the integration : $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$

$$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$

$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$

I am stuck. Please help me….

Solutions Collecting From Web of "Evaluate the integration : $\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$"

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 $$\displaystyle I = \sqrt{3}\cdot \ln\left|\frac{\sqrt{2-\sin x}-\sqrt{3}\cdot \sqrt{2+\sin x}}{\sqrt{2-\sin x}-\sqrt{3}\cdot \sqrt{2+\sin x}}\right|+2\tan^{-1}\left(\sqrt{\frac{2-\sin x}{2+\sin x}}\right)+\mathcal{C}$$

Hint:

$$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx$$

$$=\int\frac{\cos x\sqrt{4-\sin^2 x}}{(1+\sin x)(2+\sin x)}\,dx$$

( multiplying numerator & denominator by $(1+\sin x)(2+\sin x)$ under square root sign.)

Now put , $\sin x=z$.

Expand Hint :

Then ,

$$=\int\frac{1}{1+z}\sqrt{\frac{2-z}{2+z}}\,dz$$

$$=\int u\sqrt{\frac{3u-1}{u+1}}\,du\text{ , by putting $1+z=\frac{1}{u}.$ }$$

$$=\int\frac{u(3u-1)}{\sqrt{(u+1)(3u-1)}}\,du$$

$$=\int\sqrt{3u^2+2u-1}\,du-\frac{1}{2}\int \frac{d(3u^2+2u-1)}{\sqrt{3u^2+2u-1}}+2\int\frac{\,du}{\sqrt{3u^2+2u-1}}$$

$$=\cdots \cdots \cdots \cdots \cdots $$