Question

$\int \frac{1}{\sec x+\mathrm{cosec}x}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{1}{\sec x+\mathrm{cosec}x}dx \\ I=\int \frac{1}{{\displaystyle \frac{1}{\cos x}+\frac{1}{\sin x}}}dx \\ I=\frac{1}{2}\int \frac{2\sin x\cos x}{{\displaystyle \sin x+\cos x}}dx \\ I=\frac{1}{2}\int \frac{1+2\sin x\cos x-1}{{\displaystyle \sin x+\cos x}}dx \\ I=\frac{1}{2}\int \frac{{\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x-1}{{\displaystyle \sin x+\cos x}}dx \\ I=\frac{1}{2}\int \frac{{\left(\sin x+\cos x\right)}^2-1}{{\displaystyle \sin x+\cos x}}dx \\ I=\frac{1}{2}\int \frac{{\left(\sin x+\cos x\right)}^2}{{\displaystyle \sin x+\cos x}}dx-\frac{1}{2}\int \frac{1}{{\displaystyle \sin x+\cos x}}dx \\ I=\frac{1}{2}\int \left(\sin x+\cos x\right)dx-\frac{1}{2}\int \frac{1}{{\displaystyle \sin x+\cos x}}dx \\ I=\frac{1}{2}\left(-\cos x+\sin x\right)+C_1-\frac{1}{2\sqrt{2}}\int \frac{1}{{\displaystyle \frac{1}{\sqrt{2}}\left(\sin x+\cos x\right)}}dx \\ I=\frac{1}{2}\left(-\cos x+\sin x\right)+C_1-\frac{1}{2\sqrt{2}}\int \frac{1}{{\displaystyle \sin x\cos \frac{\mathrm{\pi }}{4}+\cos x\sin \frac{\mathrm{\pi }}{4}}}dx \\ I=\frac{1}{2}\left(-\cos x+\sin x\right)+C_1-\frac{1}{2\sqrt{2}}\int \frac{1}{{\displaystyle \sin \left(x+\frac{\mathrm{\pi }}{4}\right)}}dx \\ I=\frac{1}{2}\left(-\cos x+\sin x\right)+C_1-\frac{1}{2\sqrt{2}}\int \mathrm{cosec}\left(x+\frac{\mathrm{\pi }}{4}\right)dx \\ I=\frac{1}{2}\left(-\cos x+\sin x\right)-\frac{1}{2\sqrt{2}}\log \left|\tan \left(\frac{x}{2}+\frac{\mathrm{\pi }}{8}\right)\right|+C \end{gathered}$$