निश्चित समाकलन int_{frac{pi}{6}}^{frac{pi}{3} | vedclass

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Question

(C) माना $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{	an x}}$.गुणधर्म $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ का उपयोग करते

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(C) माना $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{	an x}}$.गुणधर्म $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ का उपयोग करते हुए,जहाँ $a = \frac{\pi}{6}$ और $b = \frac{\pi}{3}$,हमें $a+b = \frac{\pi}{2}$ प्राप्त होता है।$I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{	an(\frac{\pi}{2}-x)}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\sqrt{\cot x}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1+\frac{1}{\sqrt{	an x}}} = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{	an x}}{\sqrt{	an x}+1} dx$.$I$ के दोनों व्यंजकों को जोड़ने पर:$2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \left( \frac{1}{1+\sqrt{	an x}} + \frac{\sqrt{	an x}}{1+\sqrt{	an x}} ight) dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$.$2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$.अतः,$I = \frac{\pi}{12}$.

निश्चित समाकलन $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}}$ का मान ज्ञात कीजिए।

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