∫dxsin⁡x−cos⁡x+2 is equal to

# $\int \frac{\mathrm{dx}}{\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}+\sqrt{2}}$ is equal to

1. A

$-\frac{1}{\sqrt{2}}\mathrm{tan}\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)+\mathrm{C}$

2. B

$\frac{1}{\sqrt{2}}\mathrm{tan}\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)+\mathrm{C}$

3. C

$\frac{1}{\sqrt{2}}\mathrm{cot}\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)+\mathrm{C}$

4. D

$-\frac{1}{\sqrt{2}}\mathrm{cot}\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)+\mathrm{C}$

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### Solution:

$\begin{array}{l}\int \frac{\mathrm{dx}}{\mathrm{sin}\mathrm{x}-\mathrm{cos}\mathrm{x}+\sqrt{2}}\\ =\int \frac{\mathrm{dx}}{\mathrm{sin}\mathrm{x}\frac{\sqrt{2}}{\sqrt{2}}-\mathrm{cos}\mathrm{x}\frac{\sqrt{2}}{\sqrt{2}}+\sqrt{2}}\\ =\int \frac{\mathrm{dx}}{\sqrt{2}\left(\mathrm{sin}\mathrm{xsin}\frac{\mathrm{\pi }}{4}-\mathrm{cos}\mathrm{xcos}\frac{\mathrm{\pi }}{4}+1\right)}\\ =\frac{1}{\sqrt{2}}\int \frac{\mathrm{dx}}{1-\mathrm{cos}\left(\mathrm{x}+\frac{\mathrm{\pi }}{4}\right)}\\ =\frac{1}{\sqrt{2}}\int \frac{\mathrm{dx}}{1-\mathrm{cos}2\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)}=\frac{1}{\sqrt{2}}\int \frac{\mathrm{dx}}{2{\mathrm{sin}}^{2}\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)}\\ =\frac{1}{2\sqrt{2}}\int {\mathrm{cosec}}^{2}\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)\mathrm{dx}\\ =-\frac{1}{2\sqrt{2}}\cdot \frac{-\mathrm{cot}\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)}{\frac{1}{2}}+\mathrm{C}=\frac{1}{\sqrt{2}}\mathrm{cot}\left(\frac{\mathrm{x}}{2}+\frac{\mathrm{\pi }}{8}\right)+\mathrm{C}\end{array}$

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