int sin^{-1}left(sqrt{frac{x}{a+x}}right) dx | vedclass

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(D) ધારો કે $I = \int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}ight) dx$. $x = a 	an^2 	heta$ લેતા,$dx = 2a 	an 	heta \sec^2 	heta d	heta$ મળે. તેથી

Vedclass · Accepted Answer

$(a+x) 	an^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + c$

Vedclass · Answer

(D) ધારો કે $I = \int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}ight) dx$. $x = a 	an^2 	heta$ લેતા,$dx = 2a 	an 	heta \sec^2 	heta d	heta$ મળે. તેથી $\sqrt{\frac{x}{a+x}} = \sqrt{\frac{a 	an^2 	heta}{a(1 + 	an^2 	heta)}} = \sin 	heta$. માટે,$I = \int 	heta (2a 	an 	heta \sec^2 	heta) d	heta = 2a \int 	heta 	an 	heta \sec^2 	heta d	heta$. ખંડશઃ સંકલનનો ઉપયોગ કરતા,$u = 	heta$ અને $dv = 	an 	heta \sec^2 	heta d	heta$ લેતા. તેથી $du = d	heta$ અને $v = \frac{	an^2 	heta}{2}$ મળે. $I = 2a \left[ 	heta \frac{	an^2 	heta}{2} - \int \frac{	an^2 	heta}{2} d	heta ight] = a 	heta 	an^2 	heta - a \int (\sec^2 	heta - 1) d	heta$. $I = a 	heta 	an^2 	heta - a(	an 	heta - 	heta) + c = a 	heta (	an^2 	heta + 1) - a 	an 	heta + c$. અહીં $	an^2 	heta = \frac{x}{a}$ હોવાથી,$	heta = 	an^{-1} \sqrt{\frac{x}{a}}$ મળે. $I = a 	an^{-1} \sqrt{\frac{x}{a}} (\frac{x}{a} + 1) - a \sqrt{\frac{x}{a}} + c = (x+a) 	an^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + c$.

$\int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}\right) dx =$

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