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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $I = \int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}ight) dx$. Put $x = a 	an^2 	heta$,then $dx = 2a 	an 	heta \sec^2 	heta d	heta$. Then $\s

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Let $I = \int \sin^{-1}\left(\sqrt{\frac{x}{a+x}}ight) dx$. Put $x = a 	an^2 	heta$,then $dx = 2a 	an 	heta \sec^2 	heta d	heta$. Then $\sqrt{\frac{x}{a+x}} = \sqrt{\frac{a 	an^2 	heta}{a(1 + 	an^2 	heta)}} = \sqrt{\sin^2 	heta} = \sin 	heta$. So,$I = \int 	heta (2a 	an 	heta \sec^2 	heta) d	heta = 2a \int 	heta 	an 	heta \sec^2 	heta d	heta$. Using integration by parts,let $u = 	heta$ and $dv = 	an 	heta \sec^2 	heta d	heta$. Then $du = d	heta$ and $v = \frac{	an^2 	heta}{2}$. $I = 2a \left[ 	heta \frac{	an^2 	heta}{2} - \int \frac{	an^2 	heta}{2} d	heta ight] = a 	heta 	an^2 	heta - a \int (\sec^2 	heta - 1) d	heta$. $I = a 	heta 	an^2 	heta - a(	an 	heta - 	heta) + c = a 	heta (	an^2 	heta + 1) - a 	an 	heta + c$. Since $	an^2 	heta = \frac{x}{a}$,we have $	heta = 	an^{-1} \sqrt{\frac{x}{a}}$. $I = a 	an^{-1} \sqrt{\frac{x}{a}} (\frac{x}{a} + 1) - a \sqrt{\frac{x}{a}} + c = (x+a) 	an^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + c$.

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

Similar Questions

If $I = \int e^x \sin 2x \, dx$,then for what value of $K$ is $KI = e^x(\sin 2x - 2\cos 2x) + C$?

$\int \frac{\log \left(x^2+a^2\right)}{x^2} \,d x=$

If $\int {\ln ({x^2} + x)dx = x\ln ({x^2} + x) + A}$,then $A = $

$\int \frac{x \cdot \log x}{\left(\sqrt{x^2-1}\right)^3} d x=$

Integrate the function: $x \log(2x)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module