int_0^{frac{pi}{2}} frac{sin^2 x}{sin x + cos | vedclass

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Question

(B) माना $I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx$ ...$(i)$गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर:$I

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}} \log (\sqrt{2}+1)$

Vedclass · Answer

(B) माना $I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx$ ...$(i)$गुणधर्म $\int_0^a f(x) dx = \int_0^a f(a-x) dx$ का उपयोग करने पर:$I = \int_0^{\frac{\pi}{2}} \frac{\cos^2 x}{\cos x + \sin x} dx$ ...(ii)$(i)$ और (ii) को जोड़ने पर:$2I = \int_0^{\frac{\pi}{2}} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} dx = \int_0^{\frac{\pi}{2}} \frac{1}{\sin x + \cos x} dx$$2I = \frac{1}{\sqrt{2}} \int_0^{\frac{\pi}{2}} \operatorname{cosec}\left(x + \frac{\pi}{4}\right) dx$$2I = \frac{1}{\sqrt{2}} \left[ \log \left| \operatorname{cosec} \left( x + \frac{\pi}{4} \right) - \cot \left( x + \frac{\pi}{4} \right) \right| \right]_0^{\frac{\pi}{2}}$$2I = \frac{1}{\sqrt{2}} [ \log |\operatorname{cosec} \frac{3\pi}{4} - \cot \frac{3\pi}{4}| - \log |\operatorname{cosec} \frac{\pi}{4} - \cot \frac{\pi}{4}| ]$$2I = \frac{1}{\sqrt{2}} [ \log (\sqrt{2}+1) - \log (\sqrt{2}-1) ] = \frac{1}{\sqrt{2}} [ 2 \log (\sqrt{2}+1) ]$अतः,$I = \frac{1}{\sqrt{2}} \log (\sqrt{2}+1)$.

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx =$

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