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Mathematics Question on Integrals of Some Particular Functions

∫dxsin⁡x−cos⁡x+2\int \frac{dx}{\sin x - \cos x + \sqrt{2}} equals to

A

−12tan⁡(x2+π8)+C- \frac{1}{\sqrt{2}} \tan \left(\frac{x}{2} + \frac{\pi}{8}\right) + C

B

12tan⁡(x2+π8)+C\frac{1}{2} \tan \left(\frac{x}{2} + \frac{\pi}{8}\right) + C

C

12cot⁡(x2+π8)+C\frac{1}{\sqrt{2}} \cot\left(\frac{x}{2} + \frac{\pi}{8}\right) + C

D

−12cot⁡(x2+π8)+C- \frac{1}{\sqrt{2}} \cot \left(\frac{x}{2} + \frac{\pi}{8}\right) + C

Answer

−12cot⁡(x2+π8)+C- \frac{1}{\sqrt{2}} \cot \left(\frac{x}{2} + \frac{\pi}{8}\right) + C

Explanation

Solution

Let I=∫dxsin⁡x−cos⁡x+2I = \int \frac{dx}{\sin x - \cos x + \sqrt{2}}
=∫dx2(sin⁡xsin⁡π4−cos⁡xcos⁡π4+1)= \int \frac{dx}{\sqrt{2}\left(\sin x \sin \frac{\pi}{4} - \cos x \cos \frac{\pi}{4} + 1\right)}
=12∫dx1−cos⁡(x+π4)= \frac{1}{\sqrt{2}}\int \frac{dx}{1-\cos\left(x + \frac{\pi}{4}\right)}
=12∫dx2sin⁡2(x2+π8)= \frac{1}{\sqrt{2}}\int \frac{dx}{2 \sin^{2} \left(\frac{x}{2} + \frac{\pi }{8}\right)}
=122∫cosec2(x2+π8)dx= \frac{1}{2\sqrt{2}}\int cosec^{2}\left(\frac{x}{2} + \frac{\pi }{8}\right)dx
=122−cot⁡(x2+π8)12+C= \frac{1}{2\sqrt{2}} \frac{-\cot \left(\frac{x}{2}+ \frac{\pi }{8}\right)}{\frac{1}{2}} +C
=−12cot⁡(x2+π8)+C= -\frac{1}{\sqrt{2}}\cot\left(\frac{x}{2}+\frac{\pi}{8}\right)+C