Question

$\int \frac{dx}{(\sin x-\cos x+\sqrt{2})}=$

• A. $\left(\frac{-1}{\sqrt{2}}\right)\tan \left[\left(\frac{x}{2}\right)+\left(\frac{\pi }{8}\right)\right]+c$
• B. $\left(\frac{1}{\sqrt{2}}\right)\tan \left[\left(\frac{x}{2}\right)+\left(\frac{\pi }{8}\right)\right]+c$
• C. $\left(\frac{1}{\sqrt{2}}\right)\text{cot}\left[\left(\frac{x}{2}\right)+\left(\frac{\pi }{8}\right)\right]+c$
• D. $\left(\frac{-1}{\sqrt{2}}\right)\text{cot}\left[\left(\frac{x}{2}\right)+\left(\frac{\pi }{8}\right)\right]+c$

Answer 1

The correct option is D

$\left(\frac{-1}{\sqrt{2}}\right)\text{cot}\left[\left(\frac{x}{2}\right)+\left(\frac{\pi }{8}\right)\right]+c$

Explanation for the correct option:

Evaluating the integral:

$\begin{array}{l}\begin{array}{c}\mathrm{Let}I=\int \frac{dx}{\sin x-\cos x+\sqrt{2}}\\ =\int \frac{dx}{\sqrt{2}\left(\frac{1}{\sqrt{2}}\sin x-\frac{1}{\sqrt{2}}\cos x\right)+\sqrt{2}}\mathbf{[}\mathbf{\text{Multiply and divide sin⁡x−cos⁡x by}}\mathbf{}\sqrt{\mathbf{2}}\mathbf{]}\\ =\frac{1}{\sqrt{2}}\int \frac{dx}{-\cos \left(x+\frac{\pi }{4}\right)+1}\\ =\frac{1}{\sqrt{2}}\int \frac{dx}{2{\sin }^2\left(\frac{x}{2}+\frac{\pi }{8}\right)}\mathbf{[}\mathbf{cos}\left(2x\right)\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}{\mathbf{sin}}^{\mathbf{2}}\left(x\right)\mathbf{]}\\ =\frac{1}{2\sqrt{2}}\int {\mathrm{cosec}}^2\left(\frac{x}{2}+\frac{\pi }{8}\right)\mathbf{}\mathbf{[}\frac{\mathbf{1}}{\mathbf{sinx}}\mathbf{=}\mathbf{cosecx}\mathbf{]}\\ =\frac{1}{2\sqrt{2}}\int \frac{-\cot \left(\frac{x}{2}+\frac{\pi }{8}\right)}{\frac{1}{2}}+C\\ =-\frac{1}{\sqrt{2}}\cot \left(\frac{x}{2}+\frac{\pi }{8}\right)+C\end{array}\end{array}$

Hence, option (D) is the correct answer.