int_{pi / 11}^{9 pi / 22} frac{d x}{1+sqrt{ta | vedclass

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(D) ધારો કે $I = \int_{\pi/11}^{9\pi/22} \frac{dx}{1+\sqrt{	an x}}$.ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા,જ્યાં $a

Vedclass · Accepted Answer

$7 \pi / 44$

Vedclass · Answer

(D) ધારો કે $I = \int_{\pi/11}^{9\pi/22} \frac{dx}{1+\sqrt{	an x}}$.ગુણધર્મ $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ નો ઉપયોગ કરતા,જ્યાં $a = \frac{\pi}{11}$ અને $b = \frac{9\pi}{22}$,આપણને $a+b = \frac{2\pi + 9\pi}{22} = \frac{11\pi}{22} = \frac{\pi}{2}$ મળે છે.તેથી,$I = \int_{\pi/11}^{9\pi/22} \frac{dx}{1+\sqrt{	an(\pi/2 - x)}} = \int_{\pi/11}^{9\pi/22} \frac{dx}{1+\sqrt{\cot x}} = \int_{\pi/11}^{9\pi/22} \frac{\sqrt{	an x} dx}{\sqrt{	an x} + 1}$.$I$ માટેના બંને પદોનો સરવાળો કરતા:$2I = \int_{\pi/11}^{9\pi/22} \left( \frac{1}{1+\sqrt{	an x}} + \frac{\sqrt{	an x}}{1+\sqrt{	an x}} ight) dx = \int_{\pi/11}^{9\pi/22} 1 dx$.$2I = [x]_{\pi/11}^{9\pi/22} = \frac{9\pi}{22} - \frac{2\pi}{22} = \frac{7\pi}{22}$.તેથી,$I = \frac{7\pi}{44}$.

$\int_{\pi / 11}^{9 \pi / 22} \frac{d x}{1+\sqrt{\tan x}} = $

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