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

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(C) माना $I = \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx$.हम जानते हैं कि $\sin 2x = (\sin x + \cos x)^2 - 1$.अतः,$I = \int \frac{\sin x - \cos

Vedclass · Accepted Answer

$-\log |\sin x+\cos x+\sqrt{\sin 2 x}|+C$

Vedclass · Answer

(C) माना $I = \int \frac{\sin x - \cos x}{\sqrt{\sin 2x}} \, dx$.हम जानते हैं कि $\sin 2x = (\sin x + \cos x)^2 - 1$.अतः,$I = \int \frac{\sin x - \cos x}{\sqrt{(\sin x + \cos x)^2 - 1}} \, dx$.माना $t = \sin x + \cos x$.तब $dt = (\cos x - \sin x) \, dx$,जिसका अर्थ है कि $(\sin x - \cos x) \, dx = -dt$.इन मानों को समाकलन में रखने पर:$I = \int \frac{-dt}{\sqrt{t^2 - 1}} = -\int \frac{dt}{\sqrt{t^2 - 1}}$.मानक समाकलन सूत्र $\int \frac{dx}{\sqrt{x^2 - a^2}} = \log |x + \sqrt{x^2 - a^2}| + C$ का उपयोग करने पर:$I = -\log |t + \sqrt{t^2 - 1}| + C$.$t = \sin x + \cos x$ वापस रखने पर:$I = -\log |\sin x + \cos x + \sqrt{(\sin x + \cos x)^2 - 1}| + C$.चूंकि $(\sin x + \cos x)^2 - 1 = \sin 2x$,इसलिए:$I = -\log |\sin x + \cos x + \sqrt{\sin 2x}| + C$.

$\int \frac{\sin x-\cos x}{\sqrt{\sin 2 x}} \, dx$ का मान ज्ञात कीजिए।

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