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

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(A) ધારો કે $I = \int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+x\right)}{\cos x+\sin x} dx$. અહીં $\sum_{n=0}^4 \left(\frac{n \pi}{4

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) ધારો કે $I = \int_0^{\frac{\pi}{2}} \frac{\sum_{n=0}^4 \left(\frac{n \pi}{4}+xight)}{\cos x+\sin x} dx$. અહીં $\sum_{n=0}^4 \left(\frac{n \pi}{4}+xight) = (0+1+2+3+4) \frac{\pi}{4} + (1+1+1+1+1)x = \frac{10 \pi}{4} + 5x = \frac{5 \pi}{2} + 5x = 5 \left(\frac{\pi}{2} + xight)$. તેથી,$I = \int_0^{\frac{\pi}{2}} \frac{5 \left(\frac{\pi}{2} + xight)}{\sqrt{2} \left(\frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin xight)} dx = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} + xight) \sec \left(x - \frac{\pi}{4}ight) dx$ ...$(i)$. ગુણધર્મ $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ નો ઉપયોગ કરતા,$x$ ને $\left(\frac{\pi}{2} - xight)$ વડે બદલતા: $I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left[\frac{\pi}{2} + \left(\frac{\pi}{2} - xight)ight] \sec \left(\left(\frac{\pi}{2} - xight) - \frac{\pi}{4}ight) dx = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (\pi - x) \sec \left(\frac{\pi}{4} - xight) dx$. કેમ કે $\sec(\alpha) = \sec(-\alpha)$,તેથી $\sec \left(\frac{\pi}{4} - xight) = \sec \left(x - \frac{\pi}{4}ight)$. આમ,$I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} (\pi - x) \sec \left(x - \frac{\pi}{4}ight) dx$ ...(ii). $(i)$ અને (ii) નો સરવાળો કરતા: $2I = \frac{5}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \left(\frac{\pi}{2} + x + \pi - xight) \sec \left(x - \frac{\pi}{4}ight) dx = \frac{5}{\sqrt{2}} \cdot \frac{3 \pi}{2} \int_0^{\frac{\pi}{2}} \sec \left(x - \frac{\pi}{4}ight) dx$. $2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log \left| \sec \left(x - \frac{\pi}{4}ight) + 	an \left(x - \frac{\pi}{4}ight) ight| ight]_0^{\frac{\pi}{2}}$. $2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log |\sec \frac{\pi}{4} + 	an \frac{\pi}{4}| - \log |\sec \left(-\frac{\pi}{4}ight) + 	an \left(-\frac{\pi}{4}ight)| ight]$. $2I = \frac{15 \pi}{2 \sqrt{2}} \left[ \log |\sqrt{2} + 1| - \log |\sqrt{2} - 1| ight] = \frac{15 \pi}{2 \sqrt{2}} \log \left| \frac{\sqrt{2} + 1}{\sqrt{2} - 1} ight| = \frac{15 \pi}{2 \sqrt{2}} \log |(\sqrt{2} + 1)^2| = \frac{15 \pi}{\sqrt{2}} \log |\sqrt{2} + 1|$. તેથી,$I = \frac{15 \pi}{2 \sqrt{2}} \log |\sqrt{2} + 1|$.

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

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