समाकलन का मान ज्ञात कीजिए: int_0^pi left(cos^ | vedclass

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(C) माना $I = \int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$. सर्वसम

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$\sqrt{2}$

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(C) माना $I = \int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$. सर्वसमिका $\cos^2 A - \cos^2 B = \sin(B-A) \sin(B+A)$ का उपयोग करने पर: $A = \frac{3\pi}{8} - \frac{x}{4}$ और $B = \frac{11\pi}{8} + \frac{x}{4}$. $B - A = \left(\frac{11\pi}{8} + \frac{x}{4}\right) - \left(\frac{3\pi}{8} - \frac{x}{4}\right) = \frac{8\pi}{8} + \frac{2x}{4} = \pi + \frac{x}{2}$. $B + A = \left(\frac{11\pi}{8} + \frac{x}{4}\right) + \left(\frac{3\pi}{8} - \frac{x}{4}\right) = \frac{14\pi}{8} = \frac{7\pi}{4}$. अतः,समाकल्य $\sin(\pi + \frac{x}{2}) \sin(\frac{7\pi}{4}) = (-\sin \frac{x}{2}) \cdot (-\frac{1}{\sqrt{2}}) = \frac{1}{\sqrt{2}} \sin \frac{x}{2}$ हो जाता है। $I = \int_0^\pi \frac{1}{\sqrt{2}} \sin \frac{x}{2} dx = \frac{1}{\sqrt{2}} \left[ -2 \cos \frac{x}{2} \right]_0^\pi$. $I = \frac{-2}{\sqrt{2}} (\cos \frac{\pi}{2} - \cos 0) = -\sqrt{2} (0 - 1) = \sqrt{2}$.

समाकलन का मान ज्ञात कीजिए: $\int_0^\pi \left(\cos^2 \left(\frac{3\pi}{8} - \frac{x}{4}\right) - \cos^2 \left(\frac{11\pi}{8} + \frac{x}{4}\right)\right) dx$

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