Question

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

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\sin x}{\cos 2x}dx \\ =\int \left(\frac{\sin x}{2{\cos }^{2}x-1}\right)dx\left[\because \cos 2x=2{\cos }^{2}x-1\right] \\ \mathrm{Putting}\cos x=t \\ \Rightarrow -\sin xdx=dt \\ \Rightarrow \sin xdx=-dt \\ \therefore I=\int \frac{-dt}{2t^2-1} \\ =\frac{1}{2}\int \frac{-dt}{t^2-{\displaystyle \frac{1}{2}}} \\ =\frac{-1}{2}\int \frac{dt}{t^2-{\left({\displaystyle \frac{1}{\sqrt{2}}}\right)}^2} \\ =\frac{-1}{2}\times \frac{1}{2\times {\displaystyle \frac{1}{\sqrt{2}}}}\ln \left|\frac{t-{\displaystyle \frac{1}{\sqrt{2}}}}{t+{\displaystyle \frac{1}{\sqrt{2}}}}\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \int \frac{\mathit{1}}{x^{\mathit{2}}\mathit{-}{a}^{\mathit{2}}}=\frac{1}{2a}\ln \left|\frac{x\mathit{-}a}{x+a}\right|+C\right] \\ =-\frac{1}{2\sqrt{2}}\ln \left|\frac{\sqrt{2}t-1}{\sqrt{2}t+1}\right|+C \\ =-\frac{1}{2\sqrt{2}}\ln \left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right|+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\mathit{\because }t=\cos \mathit{}x\right] \\ =\frac{1}{2\sqrt{2}}\ln \left|\frac{\sqrt{2}\cos x+1}{\sqrt{2}\cos x-1}\right|+C \end{gathered}$$