int {frac{{dx}}{{sin x - cos x + sqrt 2 }}} | vedclass

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Question

(D) Let $I = \int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $.We can rewrite the denominator as $\sqrt{2} (\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqr

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $.We can rewrite the denominator as $\sqrt{2} (\frac{1}{\sqrt{2}} \sin x - \frac{1}{\sqrt{2}} \cos x + 1) = \sqrt{2} (\sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} + 1) = \sqrt{2} (\sin(x - \frac{\pi}{4}) + 1)$.Alternatively,using $\sin x - \cos x = \sqrt{2} \sin(x - \frac{\pi}{4})$,the integral becomes $I = \int \frac{dx}{\sqrt{2} \sin(x - \frac{\pi}{4}) + \sqrt{2}} = \frac{1}{\sqrt{2}} \int \frac{dx}{1 + \sin(x - \frac{\pi}{4})}$.Using the identity $1 + \sin 	heta = 1 + \cos(\frac{\pi}{2} - 	heta) = 2 \cos^2(\frac{\pi}{4} - \frac{	heta}{2})$,we have $1 + \sin(x - \frac{\pi}{4}) = 2 \cos^2(\frac{\pi}{4} - \frac{x - \pi/4}{2}) = 2 \cos^2(\frac{\pi}{4} - \frac{x}{2} + \frac{\pi}{8}) = 2 \cos^2(\frac{3\pi}{8} - \frac{x}{2})$.However,the standard approach is $I = \frac{1}{\sqrt{2}} \int \frac{dx}{1 - \cos(x + \frac{\pi}{4})} = \frac{1}{\sqrt{2}} \int \frac{dx}{2 \sin^2(\frac{x}{2} + \frac{\pi}{8})} = \frac{1}{2\sqrt{2}} \int \csc^2(\frac{x}{2} + \frac{\pi}{8}) dx$.Integrating $\csc^2(u)$ gives $-\cot(u)$,and adjusting for the coefficient $\frac{1}{2}$ of $x$,we get $I = \frac{1}{2\sqrt{2}} \cdot \frac{-\cot(\frac{x}{2} + \frac{\pi}{8})}{1/2} + c = -\frac{1}{\sqrt{2}} \cot(\frac{x}{2} + \frac{\pi}{8}) + c$.

$\int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $ equals

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