evaluate....

\hspace{-16}\int\frac{\sqrt{\cos 2x}}{\sin x}dx\\\\\\ \int\frac{\sqrt{\cos^2 x-\sin^2 x}}{\sin x}dx$\\\\\\ $\int\sqrt{\cot^2 x-1}dx$\\\\\\ Let $\cot x=t\Leftrightarrow -\csc^2 xdx=dt$\\\\ $\Leftrightarrow dx=-\frac{1}{1+t^2}dt$\\\\ $-\int\frac{\sqrt{t^2-1}}{1+t^2}dt=-\int\frac{(t^2-1)}{(1+t^2).\sqrt{t^2-1}}$\\\\\\ $-\int\frac{(1+t^2)-2}{(1+t^2).\sqrt{t^2-1}}dt$\\\\\\ $2.\int\frac{1}{(1+t^2).\sqrt{t^2-1}}dt-\int\frac{1}{\sqrt{t^2-1}}dt$\\\\\\ Now Let $I_{1}=2.\int\frac{1}{(1+t^2).\sqrt{t^2-1}}dt$ and $I_{2}=\int\frac{1}{\sqrt{t^2-1}}dt$\\\\\\ $I_{1}=2.\int\frac{1}{(1+t^2).\sqrt{t^2-1}}dt$\\\\\\ Put $t=\frac{1}{u}\Leftrightarrow dt=-\frac{1}{u^2}du$\\\\\\

\hspace{-16}-2.\int\frac{u}{(u^2+1).\sqrt{1-u^2}}du$\\\\\\ Again Put $1-u^2=v^2\Leftrightarrow du=-vdv$\\\\\\ $2.\int\frac{dv}{(\sqrt{2})^2-v^2}=\frac{2}{2.\sqrt{2}}.\ln \left|\frac{v+\sqrt{2}}{v-\sqrt{2}}\right|+C_{1}$\\\\\\ So $I_{1}=\frac{1}{\sqrt{2}}.\ln \left|\frac{\sqrt{1-u^2}+\sqrt{2}}{\sqrt{1-u^2}-\sqrt{2}}\right|+C_{1}$\\\\\\ $I_{1}=\frac{1}{\sqrt{2}}.\ln \left|\frac{\sqrt{t^2-1}+\sqrt{2}t}{\sqrt{t^2-1}+\sqrt{2}t}\right|+C_{1}$\\\\\\ $I_{1}=\frac{1}{\sqrt{2}}.\ln \left|\frac{\sqrt{\cot^2 t-1}+\sqrt{2}\cot t}{\sqrt{\cot^2 t-1}+\sqrt{2}\cot t}\right|+C_{1}$\\\\\\ Similarly $I_{2}=\frac{1}{\sqrt{t^2-1}}$\\\\\\ $=\ln \left|t+\sqrt{t^2-1}\right|+C_{2}$\\\\\\ So $I_{2}=\ln \left|\cot t+\sqrt{\cot^2 t-1}\right|+C_{2}$\\\\\\ So\\\\ $\int\frac{\sqrt{cos 2x}}{\sin x}dx =\frac{1}{\sqrt{2}}.\ln \left|\frac{\sqrt{\cot^2 t-1}+\sqrt{2}\cot t}{\sqrt{\cot^2 t-1}+\sqrt{2}\cot t}\right|+\ln \left|\cot t+\sqrt{\cot^2 t-1}\right|+C$

another solution::

\hspace{-16}{\int\frac{\sqrt{\cos2\;x}}{\sin\;x}dx}\\\\\\ {\int\frac{\sqrt{\cos2\;x}\; \sin\;x}{\sin^2\;x}dx = \int\frac{\sqrt{2\cos^2\;x-1}\; \sin\;x}{1-\cos^2\;x}dx}$\\\\\\ Let ${\cos\;x = t\Leftrightarrow \sin\;xdx = -dt}$\\\\\\ ${-\int\frac{\sqrt{2t^2-1}}{1-t^2}dt = \int\frac{2t^2-1}{\left(t^2-1\right)\sqrt{2t^2-1}}dt}$\\\\\\ Again Let ${t = \frac{1}{a}\Leftrightarrow dt = -\frac{1}{a^2}da}$\\\\\\ ${-\int\frac{\left(2-a^2\right)a}{\left(1-a^2\right)\sqrt{2-a^2}\left(a^2\right)}da}$\\\\\\ Let ${2-a^2 = p^2\Leftrightarrow -ada = pdp}$\\\\\\ ${-\int\frac{p^2\left(p\right)}{\left(p^2-1\right)\left(p^2-2\right)p}dp$\\\\\\ $-\int\frac{p^2}{\left(p^2-1\right)\left(p^2-2\right)}dp}$\\\\\\ ${\int\frac{1}{p^2-1}dp-2\int\frac{1}{p^2-\left(\sqrt{2}\right)^2}dp}$\\\\\\ ${\frac{1}{2}\ln\left|\frac{p-1}{p+1}\right|-\frac{2}{2\sqrt{2}}\ln\left|\frac{p-\sqrt{2}}{p+\sqrt{2}}\right|+C}$

Nice jagdish.

My approach (its written with some steps with trig manipulation skipped):

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

\int \frac{\sqrt{\tan \left(\frac{\pi}{4}+x \right)}}{\tan \left(\frac{\pi}{4}+x \right)-1} \ dx = \int \frac{\sqrt{\tan y}}{\tan y-1} \ dy

Now let \sqrt{\tan y}=z

Then the integral becomes

\int \frac{2z^2}{(z^2-1)(1+z^4)} \ dz = \int \frac{2z^2}{(z^2-1)(1+z^4)} \ dz

\int \frac{(z^2-1)^2-(z^4+1)}{(z^2-1)(1+z^4)} \ dz = \int \frac{(z^2-1)}{1+z^4} - \frac{1}{z^2-1}\ dz

For the 1st one, divide by z² and use the substitution

z + \frac{1}{z} =t

Second integral is well known and easy

very nice Solution bhatt sir

after that we can easily calculate