int_0^1 sqrt{frac{1-x}{1+x}} , dx is equal to | vedclass

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

Question

(A) Let $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$. Rationalizing the integrand,we get: $I = \int_0^1 \sqrt{\frac{1-x}{1+x} 	imes \frac{1-x}{1-x}} \

Vedclass · Accepted Answer

$\frac{\pi}{2}-1$

Vedclass · Answer

(A) Let $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$. Rationalizing the integrand,we get: $I = \int_0^1 \sqrt{\frac{1-x}{1+x} 	imes \frac{1-x}{1-x}} \, dx = \int_0^1 \frac{1-x}{\sqrt{1-x^2}} \, dx$. Splitting the integral: $I = \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx - \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx$. Evaluating the first part: $\int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx = [\sin^{-1} x]_0^1 = \sin^{-1}(1) - \sin^{-1}(0) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$. Evaluating the second part: Let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$. $\int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx = -\frac{1}{2} \int_1^0 u^{-1/2} \, du = \frac{1}{2} \int_0^1 u^{-1/2} \, du = \frac{1}{2} [2\sqrt{u}]_0^1 = [\sqrt{u}]_0^1 = 1 - 0 = 1$. Thus,$I = \frac{\pi}{2} - 1$.

$\int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$ is equal to

Similar Questions

If $f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \\ 4x^3 - 1, & x > 1 \end{cases}$,then $\int_{0}^{2} f(x) dx$ is

$A$ minimum value of $\int_0^x t e^{t^2} d t$ is

Evaluate $\int_{-2}^1 f(x) dx$,where $f(x) = \begin{cases} 1-2x, & x \leq 0 \\ 1+2x, & x \geq 0 \end{cases}$

$\int_0^1 \frac{x}{(1-x)^{3/4}} dx = $

$\int_0^1 |5x - 3| \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module