int sqrt{frac{2+x}{2-x}} , dx is equal to | vedclass

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

Question

(D) Let $I = \int \sqrt{\frac{2+x}{2-x}} \, dx$. Multiplying the numerator and denominator inside the square root by $\sqrt{2+x}$: $I = \int \frac{2+x

Vedclass · Accepted Answer

$2 \sin^{-1}\left(\frac{x}{2}\right) - \sqrt{4-x^2} + C$

Vedclass · Answer

(D) Let $I = \int \sqrt{\frac{2+x}{2-x}} \, dx$. Multiplying the numerator and denominator inside the square root by $\sqrt{2+x}$: $I = \int \frac{2+x}{\sqrt{(2-x)(2+x)}} \, dx = \int \frac{2+x}{\sqrt{4-x^2}} \, dx$. Splitting the integral into two parts: $I = 2 \int \frac{1}{\sqrt{2^2-x^2}} \, dx + \int \frac{x}{\sqrt{4-x^2}} \, dx$. For the first part, use the standard integral $\int \frac{1}{\sqrt{a^2-x^2}} \, dx = \sin^{-1}(\frac{x}{a})$. For the second part, let $u = 4-x^2$, so $du = -2x \, dx$ or $x \, dx = -\frac{1}{2} du$. $I = 2 \sin^{-1}\left(\frac{x}{2}\right) - \frac{1}{2} \int u^{-1/2} \, du = 2 \sin^{-1}\left(\frac{x}{2}\right) - \frac{1}{2} (2u^{1/2}) + C$. $I = 2 \sin^{-1}\left(\frac{x}{2}\right) - \sqrt{4-x^2} + C$.

$\int \sqrt{\frac{2+x}{2-x}} \, dx$ is equal to

Similar Questions

$\int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} dx =$

$\int \frac{1}{x \sqrt{x^6+1}} \, dx =$

$\int \frac{x^4+1}{x^6+1} \, dx =$

If $f(x) = \int \operatorname{cosec}^5 x \, dx$,then $f\left(\frac{\pi}{4}\right) = $

$\int \sin ^4 x \cos ^4 x \, dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module