Q105 / 200 · Manipal (MET) 2014 · Mathematics

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

  • A
    $-\frac{1}{\sqrt{2}} \tan \left(\frac{x}{2}+\frac{\pi}{8}\right)+C$
  • B
    $\frac{1}{\sqrt{2}} \tan \left(\frac{x}{2}+\frac{\pi}{8}\right)+C$
  • C
    $\frac{1}{\sqrt{2}} \cot \left(\frac{x}{2}+\frac{\pi}{8}\right)+C$
  • Correct answer
    D
    $-\frac{1}{\sqrt{2}} \cot \left(\frac{x}{2}+\frac{\pi}{8}\right)+C$

Solution

$\begin{aligned} & \text { Let, } I=\int \frac{d x}{\sin x-\cos x+\sqrt{2}} \\ & \begin{aligned} & \Rightarrow I=\int \frac{d x}{\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin x-\frac{1}{\sqrt{2}} \cos x\right)+\sqrt{2}} \\ &= \frac{1}{\sqrt{2}} \int \frac{d x}{1-\cos \left(x+\frac{\pi}{4}\right)}=\frac{1}{\sqrt{2}} \int \frac{d x}{2 \sin ^2\left(\frac{x}{2}+\frac{\pi}{8}\right)} \\ &=\frac{1}{2 \sqrt{2}} \int \frac{d x}{\sin ^2\left(\frac{x}{2}+\frac{\pi}{8}\right)}=\frac{1}{2 \sqrt{2}} \\ &= \frac{1}{2 \sqrt{2}}\left[\frac{-\cot \left(\frac{x}{2}+\frac{\pi}{8}\right)}{\frac{1}{2}}\right]+C \\ &=-\frac{1}{\sqrt{2}} \cot \left(\frac{x}{2}+\frac{\pi}{8}\right)+C\end{aligned}\end{aligned}$

Hence, the correct answer is option: 'D'.

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