int text{cosec}^{-1} left( sqrt{frac{a + x}{x | vedclass

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(B) Let $I = \int 	ext{cosec}^{-1} \left( \sqrt{\frac{a + x}{x}} ight) dx$. Since $	ext{cosec}^{-1} \sqrt{\frac{a+x}{x}} = \sin^{-1} \sqrt{\frac{x

Vedclass · Accepted Answer

$a 	heta 	an^2 	heta - a 	an 	heta + a 	heta + c$ (where $	heta = 	an^{-1} \left( \sqrt{\frac{x}{a}} ight)$)

Vedclass · Answer

(B) Let $I = \int 	ext{cosec}^{-1} \left( \sqrt{\frac{a + x}{x}} ight) dx$. Since $	ext{cosec}^{-1} \sqrt{\frac{a+x}{x}} = \sin^{-1} \sqrt{\frac{x}{a+x}}$, let $x = a 	an^2 	heta$. Then $dx = 2a 	an 	heta \sec^2 	heta d	heta$.Substituting these, $I = \int \sin^{-1} \left( \sqrt{\frac{a 	an^2 	heta}{a + a 	an^2 	heta}} ight) (2a 	an 	heta \sec^2 	heta) d	heta = \int \sin^{-1} (\sin 	heta) (2a 	an 	heta \sec^2 	heta) d	heta = \int 	heta (2a 	an 	heta \sec^2 	heta) d	heta$.Using integration by parts: $I = 2a \left[ 	heta \int 	an 	heta \sec^2 	heta d	heta - \int \left( 1 \cdot \int 	an 	heta \sec^2 	heta d	heta ight) d	heta ight] = 2a \left[ 	heta \cdot \frac{	an^2 	heta}{2} - \int \frac{	an^2 	heta}{2} d	heta ight] = a 	heta 	an^2 	heta - a \int (\sec^2 	heta - 1) d	heta$.$I = a 	heta 	an^2 	heta - a (	an 	heta - 	heta) + c = a 	heta 	an^2 	heta - a 	an 	heta + a 	heta + c$.

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

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