sqrt{2} int frac{sin x , dx}{sin left( x - fr | vedclass

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Question

(C) माना $I = \sqrt{2} \int \frac{\sin x}{\sin \left( x - \frac{\pi}{4} \right)} dx$. $x = (x - \frac{\pi}{4}) + \frac{\pi}{4}$ प्रतिस्थापित करने पर।

Vedclass · Accepted Answer

$x + \log \left| \sin \left( x - \frac{\pi}{4} \right) \right| + c$

Vedclass · Answer

(C) माना $I = \sqrt{2} \int \frac{\sin x}{\sin \left( x - \frac{\pi}{4} \right)} dx$. $x = (x - \frac{\pi}{4}) + \frac{\pi}{4}$ प्रतिस्थापित करने पर। $I = \sqrt{2} \int \frac{\sin \left( (x - \frac{\pi}{4}) + \frac{\pi}{4} \right)}{\sin \left( x - \frac{\pi}{4} \right)} dx$. $\sin(A+B) = \sin A \cos B + \cos A \sin B$ सूत्र का उपयोग करने पर: $I = \sqrt{2} \int \frac{\sin(x - \frac{\pi}{4}) \cos(\frac{\pi}{4}) + \cos(x - \frac{\pi}{4}) \sin(\frac{\pi}{4})}{\sin(x - \frac{\pi}{4})} dx$. चूंकि $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ और $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$: $I = \sqrt{2} \int \left( \frac{1}{\sqrt{2}} + \cot(x - \frac{\pi}{4}) \cdot \frac{1}{\sqrt{2}} \right) dx$. $I = \int (1 + \cot(x - \frac{\pi}{4})) dx$. $I = x + \ln \left| \sin(x - \frac{\pi}{4}) \right| + c$.

$\sqrt{2} \int \frac{\sin x \, dx}{\sin \left( x - \frac{\pi}{4} \right)} = $

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