Integral Calculus
86265
\(\int \frac{d x}{\sin x-\cos x+\sqrt{2}}\) is equal to
1 \(-\frac{1}{\sqrt{2}} \tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}\)
2 \(\frac{1}{\sqrt{2}} \tan \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}\)
3 \(\frac{1}{\sqrt{2}} \cot \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}\)
4 \(-\frac{1}{\sqrt{2}} \cot \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}\)
Explanation:
(D) : Let,
\(I=\int \frac{d x}{\sin x-\cos x+\sqrt{2}}\)
\(=\int \frac{d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x\right)+\sqrt{2}}\)
\(=\frac{1}{\sqrt{2}} \int \frac{\mathrm{dx}}{-\cos \left(\mathrm{x}+\frac{\pi}{4}\right)+1}\)
\(=\frac{1}{\sqrt{2}} \int \frac{\mathrm{dx}}{2 \sin ^{2}\left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)}\)
\(=\frac{1}{2 \sqrt{2}} \int \operatorname{cosec}^{2}\left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)\)
\(=\frac{1}{2 \sqrt{2}} \times \frac{-\cot \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)}{\frac{1}{2}}+\mathrm{C}\)
\(=-\frac{1}{\sqrt{2}} \cot \left(\frac{\mathrm{x}}{2}+\frac{\pi}{8}\right)+\mathrm{C}\)