int_{0}^{1} sin left( 2 tan^{-1} sqrt{frac{1+ | vedclass

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(B) ધારો કે $I = \int_{0}^{1} \sin \left( 2 	an^{-1} \sqrt{\frac{1+x}{1-x}} ight) \, dx$. $x = \cos 	heta$ આદેશ લેતા,$dx = -\sin 	heta \, d	heta

Vedclass · Accepted Answer

$\pi / 4$

Vedclass · Answer

(B) ધારો કે $I = \int_{0}^{1} \sin \left( 2 	an^{-1} \sqrt{\frac{1+x}{1-x}} ight) \, dx$. $x = \cos 	heta$ આદેશ લેતા,$dx = -\sin 	heta \, d	heta$ મળે. જ્યારે $x = 0$,ત્યારે $	heta = \pi / 2$ અને જ્યારે $x = 1$,ત્યારે $	heta = 0$. સાઇન વિધેયની અંદરનું પદ: $2 	an^{-1} \sqrt{\frac{1+\cos 	heta}{1-\cos 	heta}} = 2 	an^{-1} \sqrt{\frac{2 \cos^2(	heta/2)}{2 \sin^2(	heta/2)}} = 2 	an^{-1} (\cot(	heta/2)) = 2 	an^{-1} \left( 	an \left( \frac{\pi}{2} - \frac{	heta}{2} ight) ight) = 2 \left( \frac{\pi}{2} - \frac{	heta}{2} ight) = \pi - 	heta$. આમ,સંકલન નીચે મુજબ થશે: $I = \int_{\pi/2}^{0} \sin(\pi - 	heta) (- \sin 	heta) \, d	heta = \int_{0}^{\pi/2} \sin^2 	heta \, d	heta$. નિત્યસમ $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$ નો ઉપયોગ કરતા: $I = \int_{0}^{\pi/2} \frac{1 - \cos 2	heta}{2} \, d	heta = \left[ \frac{	heta}{2} - \frac{\sin 2	heta}{4} ight]_{0}^{\pi/2} = \left( \frac{\pi}{4} - 0 ight) - (0 - 0) = \frac{\pi}{4}$.

$\int_{0}^{1} \sin \left( 2 \tan^{-1} \sqrt{\frac{1+x}{1-x}} \right) \, dx = $

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