Q64 / 120 · VITEEE 2012 · Mathematics
$\int \frac{d x}{\sin x-\cos x+\sqrt{2}}$ equals to
- A$-\frac{1}{\sqrt{2}} \tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}$
- B$\frac{1}{2} \tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}$
- Correct answerC$\frac{1}{\sqrt{2}} \cot \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}$
- D$-\frac{1}{\sqrt{2}} \cot \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}$
Solution
$$
\begin{array}{l}
\text { Let } I=\int \frac{d x}{\sin x-\cos x+\sqrt{2}} \\
=\int \frac{d x}{\sqrt{2}\left(\sin x \sin \frac{\pi}{4}-\cos x \cos \frac{\pi}{4}+1\right)} \\
=\frac{1}{\sqrt{2}} \int \frac{d x}{1-\cos \left(x+\frac{\pi}{4}\right)} \\
=\frac{1}{\sqrt{2}} \int \frac{d x}{2 \sin ^{2}\left(\frac{x}{2}+\frac{\pi}{8}\right)} \\
=\frac{-1}{\sqrt{2}} \int \operatorname{cosec}^{2}\left(\frac{x}{2}+\frac{\pi}{8}\right) d x \\
=\frac{-1}{2 \sqrt{2}} \cdot \frac{-\cot \left(\frac{x}{2}+\frac{\pi}{8}\right)}{\frac{1}{2}}+C \\
\left.=\frac{x}{\sqrt{2}}+\frac{\pi}{8}\right)+C
\end{array}
$$
Hence, the correct answer is option: 'C'.