int_0^{frac{pi}{2}} frac{sum_{n=0}^4 left(fra | vedclass

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(A) माना $I = \int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right)}{\cos x+\sin x} dx$ है। चूंकि $\sum_{n=0}^4 \left(\frac{n \pi}{

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$I=\frac{15 \pi}{2 \sqrt{2}} \log |\sqrt{2}+1|$

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(A) माना $I = \int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+xight)}{\cos x+\sin x} dx$ है। चूंकि $\sum_{n=0}^4 \left(\frac{n \pi}{4}+xight) = (0+1+2+3+4) \frac{\pi}{4} + (1+1+1+1+1)x = \frac{10 \pi}{4} + 5x = \frac{5 \pi}{2} + 5x = 5 \left(\frac{\pi}{2} + xight)$ है। अतः,$I = \int_0^{\frac{\pi}{2}} \frac{5 \left(\frac{\pi}{2} + xight)}{\sqrt{2} \left(\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin xight)} dx = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} + xight) \sec \left(x - \frac{\pi}{4}ight) dx$ ...$(i)$। गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करते हुए,$x$ को $\left(\frac{\pi}{2} - xight)$ से बदलने पर: $I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left[\frac{\pi}{2} + \left(\frac{\pi}{2} - xight)ight] \sec \left(\left(\frac{\pi}{2} - xight) - \frac{\pi}{4}ight) dx = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (\pi - x) \sec \left(\frac{\pi}{4} - xight) dx$। चूंकि $\sec(\alpha) = \sec(-\alpha)$,इसलिए $\sec \left(\frac{\pi}{4} - xight) = \sec \left(x - \frac{\pi}{4}ight)$। अतः,$I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (\pi - x) \sec \left(x - \frac{\pi}{4}ight) dx$ ...(ii)। $(i)$ और (ii) को जोड़ने पर: $2I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} + x + \pi - xight) \sec \left(x - \frac{\pi}{4}ight) dx = \frac{5}{\sqrt{2}} \cdot \frac{3 \pi}{2} \int_0^{\frac{\pi}{2}} \sec \left(x - \frac{\pi}{4}ight) dx$। $2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log \left| \sec \left(x - \frac{\pi}{4}ight) + 	an \left(x - \frac{\pi}{4}ight) ight| ight]_0^{\frac{\pi}{2}}$। $2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log |\sec \frac{\pi}{4} + 	an \frac{\pi}{4}| - \log |\sec \left(-\frac{\pi}{4}ight) + 	an \left(-\frac{\pi}{4}ight)| ight]$। $2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log |\sqrt{2} + 1| - \log |\sqrt{2} - 1| ight] = \frac{15 \pi}{2 \sqrt{2}} \log \left| \frac{\sqrt{2} + 1}{\sqrt{2} - 1} ight| = \frac{15 \pi}{2 \sqrt{2}} \log |(\sqrt{2} + 1)^2| = \frac{15 \pi}{\sqrt{2}} \log |\sqrt{2} + 1|$। अतः,$I = \frac{15 \pi}{2 \sqrt{2}} \log |\sqrt{2} + 1|$।

$\int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right)}{\cos x+\sin x} d x=$

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