int sin ^{-1} sqrt{frac{x}{a+x}} d x= | vedclass

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

Question

(A) $I = \int \sin ^{-1} \sqrt{\frac{x}{a+x}} dx$ Let $x = a 	an^2 t$,then $dx = 2a 	an t \sec^2 t dt$. Substituting these into the integral: $I = \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) $I = \int \sin ^{-1} \sqrt{\frac{x}{a+x}} dx$ Let $x = a 	an^2 t$,then $dx = 2a 	an t \sec^2 t dt$. Substituting these into the integral: $I = \int \sin^{-1} \sqrt{\frac{a 	an^2 t}{a(1 + 	an^2 t)}} (2a 	an t \sec^2 t) dt$ $I = \int \sin^{-1} \sqrt{\frac{	an^2 t}{\sec^2 t}} (2a 	an t \sec^2 t) dt$ $I = \int t (2a 	an t \sec^2 t) dt = 2a \int t 	an t \sec^2 t dt$ Using integration by parts: $I = 2a \left[ t \int 	an t \sec^2 t dt - \int (1 \cdot \int 	an t \sec^2 t dt) dt ight]$ Since $\int 	an t \sec^2 t dt = \frac{	an^2 t}{2}$: $I = 2a \left[ t \cdot \frac{	an^2 t}{2} - \int \frac{	an^2 t}{2} dt ight] = a t 	an^2 t - a \int (\sec^2 t - 1) dt$ $I = a t 	an^2 t - a (	an t - t) + C = a t 	an^2 t - a 	an t + at + C$ $I = a t (	an^2 t + 1) - a 	an t + C = a t \sec^2 t - a 	an t + C$ Substituting $t = 	an^{-1} \sqrt{\frac{x}{a}}$ and $	an^2 t = \frac{x}{a}$: $I = a 	an^{-1} \sqrt{\frac{x}{a}} (1 + \frac{x}{a}) - a \sqrt{\frac{x}{a}} + C$ $I = (a+x) 	an^{-1} \sqrt{\frac{x}{a}} - \sqrt{ax} + C$

$\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x=$

Similar Questions

Integrate the function: $\int \frac{3x}{1+2x^4} dx$

$\int \frac{1}{x^3} [\log x^x]^2 \, dx = $

For which of the following functions is the substitution ${x^2} = t$ applicable?

$\int \frac{30x^{14} + 15^x \log 225}{x^{15} + 15^x} dx =$

$\int \frac{x^3}{\sqrt{1 - x^8}} \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module