int_0^1 sqrt{frac{2+x}{2-x}} , dx = | vedclass

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

Question

(D) Let $I = \int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx$. Rationalizing the integrand: $I = \int_0^1 \frac{2+x}{\sqrt{4-x^2}} \, dx = \int_0^1 \frac{2}{\sq

Vedclass · Accepted Answer

$\frac{\pi}{3}+2-\sqrt{3}$

Vedclass · Answer

(D) Let $I = \int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx$. Rationalizing the integrand: $I = \int_0^1 \frac{2+x}{\sqrt{4-x^2}} \, dx = \int_0^1 \frac{2}{\sqrt{4-x^2}} \, dx + \int_0^1 \frac{x}{\sqrt{4-x^2}} \, dx$. For the first part,$\int_0^1 \frac{2}{\sqrt{2^2-x^2}} \, dx = 2 \left[ \sin^{-1} \left( \frac{x}{2} \right) \right]_0^1 = 2 \left( \sin^{-1} \frac{1}{2} - \sin^{-1} 0 \right) = 2 \left( \frac{\pi}{6} - 0 \right) = \frac{\pi}{3}$. For the second part,let $t = 4-x^2$,then $dt = -2x \, dx$,so $x \, dx = -\frac{1}{2} \, dt$. $\int_0^1 \frac{x}{\sqrt{4-x^2}} \, dx = -\frac{1}{2} \int_4^3 \frac{1}{\sqrt{t}} \, dt = \frac{1}{2} \int_3^4 t^{-1/2} \, dt = \frac{1}{2} [2\sqrt{t}]_3^4 = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3}$. Adding both parts: $I = \frac{\pi}{3} + 2 - \sqrt{3}$.

$\int_0^1 \sqrt{\frac{2+x}{2-x}} \, dx =$

Similar Questions

If $f(x) = \begin{cases} 4x + 3, & 1 \le x \le 2 \\ 3x + 5, & 2 < x \le 4 \end{cases}$,then $\int_1^4 f(x) \, dx = $

If $\int_0^1 {x \log \left( {1 + \frac{x}{2}} \right)} \,dx = a + b \log \frac{2}{3},$ then

$\int_{-2}^{2} |1 - x^2| \, dx = $

$\int_{0}^{\frac{\pi}{2}} \frac{\sin x \cos x}{1+\sin ^{4} x} d x=$

$\int_0^\infty \frac{dx}{(x + \sqrt{x^2 + 1})^3} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module