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## $\mathrm{Let}I=\int \frac{1}{\mathrm{sin}x+\mathrm{cos}x}dx\phantom{\rule{0ex}{0ex}}\mathrm{Putting}\mathrm{sin}x=\frac{2\mathrm{tan}\left(\frac{x}{2}\right)}{1+{\mathrm{tan}}^{2}\left(\frac{x}{2}\right)}\mathrm{and}\mathrm{cos}x=\frac{1-{\mathrm{tan}}^{2}\left(\frac{x}{2}\right)}{1+{\mathrm{tan}}^{2}\left(\frac{x}{2}\right)}\phantom{\rule{0ex}{0ex}}=\int \frac{1}{\frac{2\mathrm{tan}\left(\frac{x}{2}\right)}{1+{\mathrm{tan}}^{2}\left(\frac{x}{2}\right)}+\frac{1-{\mathrm{tan}}^{2}\left(\frac{x}{2}\right)}{1+{\mathrm{tan}}^{2}\left(\frac{x}{2}\right)}}dx\phantom{\rule{0ex}{0ex}}=\int \frac{{\mathrm{sec}}^{2}\left(\frac{x}{2}\right)}{1-{\mathrm{tan}}^{2}\left(\frac{x}{2}\right)+2\mathrm{tan}\left(\frac{x}{2}\right)}dx\phantom{\rule{0ex}{0ex}}\mathrm{Let}\mathrm{tan}\left(\frac{x}{2}\right)=t\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}{\mathrm{sec}}^{2}\left(\frac{x}{2}\right)dx=dt\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sec}}^{2}\left(\frac{x}{2}\right)dx=2dt\phantom{\rule{0ex}{0ex}}\therefore I=2\int \frac{dt}{1-{t}^{2}+2t}\phantom{\rule{0ex}{0ex}}=-2\int \frac{dt}{{t}^{2}-2t-1}\phantom{\rule{0ex}{0ex}}=-2\int \frac{dt}{{t}^{2}-2t+1-2}\phantom{\rule{0ex}{0ex}}=2\int \frac{dt}{{\left(\sqrt{2}\right)}^{2}-{\left(t-1\right)}^{2}}\phantom{\rule{0ex}{0ex}}=2×\frac{1}{2\sqrt{2}}\mathrm{ln}\left|\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right|+C\phantom{\rule{0ex}{0ex}}=\frac{1}{\sqrt{2}}\mathrm{ln}\left|\frac{\sqrt{2}+\mathrm{tan}\frac{x}{2}-1}{\sqrt{2}-\mathrm{tan}\frac{x}{2}+1}\right|+C$

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