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

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Question

(D) ધારો કે $I = \int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $.છેદને $\sqrt{2} (\sin x \cdot \frac{1}{\sqrt{2}} - \cos x \cdot \frac{1}{\sqrt{2}}

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) ધારો કે $I = \int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $.છેદને $\sqrt{2} (\sin x \cdot \frac{1}{\sqrt{2}} - \cos x \cdot \frac{1}{\sqrt{2}} + 1) = \sqrt{2} (\sin x \cos \frac{\pi}{4} - \cos x \sin \frac{\pi}{4} + 1) = \sqrt{2} (1 - \cos(x + \frac{\pi}{4}))$ તરીકે લખી શકાય.તેથી,$I = \frac{1}{\sqrt{2}} \int \frac{dx}{1 - \cos(x + \frac{\pi}{4})} = \frac{1}{\sqrt{2}} \int \frac{dx}{2 \sin^2(\frac{x}{2} + \frac{\pi}{8})}$.આમ,$I = \frac{1}{2\sqrt{2}} \int \csc^2(\frac{x}{2} + \frac{\pi}{8}) dx$.સંકલન કરતા,$I = \frac{1}{2\sqrt{2}} \cdot \frac{-\cot(\frac{x}{2} + \frac{\pi}{8})}{1/2} + c = -\frac{1}{\sqrt{2}} \cot(\frac{x}{2} + \frac{\pi}{8}) + c$.

$\int {\frac{{dx}}{{\sin x - \cos x + \sqrt 2 }}} $ ની કિંમત શોધો.

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