Question

$\int \frac{\sqrt{\cos 2x}}{\sin x}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{\sqrt{\cos 2x}}{\sin x}dx \\ =\int \sqrt{\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}}\times \frac{1}{\sin x}dx \\ =\int \frac{\sqrt{1-{\tan }^{2}x}}{\sec x\sin x}dx \\ =\int \frac{\sqrt{1-{\tan }^{2}x}}{\tan x}dx \\ \mathrm{Putting}1-{\tan }^{2}x=t^2 \\ \mathrm{Diff}\mathrm{both}\mathrm{sides}\mathrm{w}.\mathrm{r}.\mathrm{t}x \\ -2\tan x{\sec }^{2}x=2t\frac{dt}{dx} \\ \Rightarrow dx=\frac{-2tdt}{2\tan x{\sec }^{2}x} \\ \therefore I=\int \frac{t\times \left(-2t\right)dt}{2{\tan }^{2}x{\sec }^{2}x} \\ =\int \frac{-t^{2}dt}{\left(1-t^2\right)\left(1+{\tan }^{2}x\right)} \\ =\int \frac{-t^{2}dt}{\left(1-t^2\right)\left(2-t^2\right)} \\ \mathrm{Let}{t}^2=p \\ \mathrm{Then}\frac{t^2}{\left(1-t^2\right)\left(2-t^2\right)}=\frac{p}{\left(1-p\right)\left(2-p\right)} \\ \mathrm{And}\mathrm{let}\frac{p}{\left(1-p\right)\left(2-p\right)}=\frac{A}{1-p}+\frac{B}{2-p} \\ \Rightarrow \frac{p}{\left(1-p\right)\left(2-p\right)}=\frac{A\left(2-p\right)+B\left(1-p\right)}{\left(1-p\right)\left(2-p\right)} \\ \Rightarrow p=A\left(2-p\right)+B\left(1-p\right) \\ \mathrm{Putting}p=2 \\ \therefore 2=A\times 0+B\left(1-2\right) \\ \Rightarrow B=-2 \\ \mathrm{Putting}p=1 \\ \therefore 1=A\left(2-1\right)+B\times 0 \\ \Rightarrow A=1 \\ \therefore \frac{p}{\left(1-p\right)\left(2-p\right)}=\frac{1}{1-p}-\frac{2}{2-p} \\ \therefore I=-\int \frac{dt}{1-t^2}+2\int \frac{dt}{2-t^2} \\ =-\int \frac{dt}{1^2-t^2}+2\int \frac{dt}{{\left(\sqrt{2}\right)}^2-t^2} \\ =-\frac{1}{2}\log \left|\frac{1+t}{1-t}\right|+2\times \frac{1}{2\sqrt{2}}\log \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+C \\ =-\frac{1}{2}\log \left|\frac{1+\sqrt{1-{\tan }^{2}x}}{1-\sqrt{1-{\tan }^{2}x}}\right|+\frac{1}{\sqrt{2}}\log \left|\frac{\sqrt{2}+\sqrt{1-{\tan }^{2}x}}{\sqrt{2}-\sqrt{1-{\tan }^{2}x}}\right|+C \\ =-\frac{1}{2}\log \left|\frac{1+\sqrt{1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}}}{1-\sqrt{1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}}}\right|+\frac{1}{\sqrt{2}}\log \left|\frac{\sqrt{2}+\sqrt{1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}}}{\sqrt{2}-\sqrt{1-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}}}\right|+C \\ =-\frac{1}{2}\log \left|\frac{\cos x+\sqrt{{\cos }^{2}x-{\sin }^{2}x}}{\cos x-\sqrt{{\cos }^{2}x-{\sin }^{2}x}}\right|+\frac{1}{\sqrt{2}}\log \left|\frac{\sqrt{2}\cos x+\sqrt{{\cos }^2-{\sin }^{2}x}}{\sqrt{2}\cos x-\sqrt{{\cos }^2-{\sin }^{2}x}}\right|+C \\ =-\frac{1}{2}\log \left|\frac{\cos x+\sqrt{\cos 2x}}{\cos x-\sqrt{\cos 2x}}\right|+\frac{1}{\sqrt{2}}\log \left|\frac{\sqrt{2}\cos x+\sqrt{\cos 2x}}{\sqrt{2}\cos x-\sqrt{\cos 2x}}\right|+C \\ =-\frac{1}{2}\log \left|\frac{\cos x+\sqrt{\cos 2x}}{\cos x-\sqrt{\cos 2x}}\right|+\frac{1}{\sqrt{2}}\log \left|\frac{\left(\sqrt{2}\cos x+\sqrt{\cos 2x}\right)\left(\sqrt{2}\cos x+\sqrt{\cos 2x}\right)}{\left(\sqrt{2}\cos x-\sqrt{\cos 2x}\right)\left(\sqrt{2}\cos x+\sqrt{\cos 2x}\right)}\right|+C \\ =\frac{1}{2}\log \left|\frac{\cos x-\sqrt{\cos 2x}}{\cos x+\sqrt{\cos 2x}}\right|+\frac{1}{\sqrt{2}}\log \left|\frac{{\left(\sqrt{2}\cos x+\sqrt{\cos 2x}\right)}^2}{\left(2{\cos }^{2}x-\cos 2x\right)}\right|+C \\ =\frac{1}{2}\log \left|\frac{\cos x-\sqrt{\cos 2x}}{\cos x+\sqrt{\cos 2x}}\right|+\frac{1}{\sqrt{2}}\log \left|\frac{{\left(\sqrt{2}\cos x+\sqrt{\cos 2x}\right)}^2}{1}\right|+C \\ =\frac{1}{2}\log \left|\frac{\cos x-\sqrt{\cos 2x}}{\cos x+\sqrt{\cos 2x}}\right|+\frac{1}{\sqrt{2}}\times 2\log \left|\sqrt{2}\cos x+\sqrt{\cos 2x}\right|+C \\ =\sqrt{2}\log \left|\sqrt{2}\cos x+\sqrt{\cos 2x}\right|+\frac{1}{2}\log \left|\frac{\cos x-\sqrt{\cos 2x}}{\cos x+\sqrt{\cos 2x}}\right|+C \end{gathered}$$