Question

Answer 1

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