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Question: \[\int_{}^{}\frac{dx}{\sin x–\cos x + \sqrt{2}} =\]...

∫dxsin⁡x–cos⁡x+2=\int_{}^{}\frac{dx}{\sin x–\cos x + \sqrt{2}} =

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}{\sqrt{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

=

=

= 1(2)∫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∫cos⁡ec2(x2+π8)dx\frac{1}{2\sqrt{2}}\int_{}^{}{\cos ec^{2}\left( \frac{x}{2} + \frac{\pi}{8} \right)}dx = –cot⁡(x2+π8)dx(22)(12)+C–\frac{\cot\left( \frac{x}{2} + \frac{\pi}{8} \right)dx}{\left( 2\sqrt{2} \right)\left( \frac{1}{2} \right)} + C

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