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$$\displaystyle \int{\frac{\sqrt{\cos 2x}}{\sin x}dx}$$ $$\displaystyle =\int{\frac{\sqrt{\frac{1-{{\tan }^{2}}x}{1+{{\tan }^{2}}x}}}{\sin x}}\,\,dx$$ $$\displaystyle =\int{\frac{\sqrt{1-{{\tan }^{2}}x}}{\sec x.\sin x}\,dx}$$ $$\displaystyle =\int{\frac{\sqrt{1-{{\tan }^{2}}x}}{\tan x}\,\,dx}$$ $$\displaystyle =\int{\frac{\sqrt{1-{{\tan }^{2}}x}}{\tan x\left( 1+{{\tan }^{2}}x \right)}\,{{\sec }^{2}}x\,\,dx}$$ $$\displaystyle =\int{\frac{\sqrt{1-{{\tan }^{2}}x}}{{{\tan }^{2}}x\left( 1+{{\tan }^{2}}x \right)}}\,\,\tan x{{\sec }^{2}}x\,\,dx$$ Let $$\displaystyle \sqrt{1-{{\tan }^{2}}x}\,\,=t$$ $$\displaystyle 1-{{\tan }^{2}}x={{t}^{2}}$$ Differentiate both sides w.r.t x $$\displaystyle -2\tan x{{\sec }^{2}}x\,\,dx\,\,=2\,tdt$$ \(\displaystyle \Rightarrow I=-\int{\frac{t}{\left( 1-{{t}^{2}} \right)\left( […]