<s>[INST] 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....
 [/INST] 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 ,
$$=\int\frac{1}{1+z}\sqrt{\frac{2-z}{2+z}}\,dz$$
$$=\int u\sqrt{\frac{3u-1}{u+1}}\,du\text{ , by putting $1+z=\frac{1}{u}.$ }$$
$$=\int\frac{u(3u-1)}{\sqrt{(u+1)(3u-1)}}\,du$$
$$=\int\sqrt{3u^2+2u-1}\,du-\frac{1}{2}\int \frac{d(3u^2+2u-1)}{\sqrt{3u^2+2u-1}}+2\int\frac{\,du}{\sqrt{3u^2+2u-1}}$$
$$=\cdots   \cdots \cdots \cdots \cdots $$
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