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

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

Question

(D) To evaluate the integral $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the integrand: $I = \int_0^1 \frac{\sqrt{1-x}}{\sqrt{1+x}} \cd

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) To evaluate the integral $I = \int_0^1 \sqrt{\frac{1-x}{1+x}} \, dx$,we rationalize the integrand: $I = \int_0^1 \frac{\sqrt{1-x}}{\sqrt{1+x}} \cdot \frac{\sqrt{1-x}}{\sqrt{1-x}} \, dx = \int_0^1 \frac{1-x}{\sqrt{1-x^2}} \, dx$ Split the integral into two parts: $I = \int_0^1 \frac{1}{\sqrt{1-x^2}} \, dx - \int_0^1 \frac{x}{\sqrt{1-x^2}} \, dx$ For the second part,let $u = 1-x^2$,then $du = -2x \, dx$,so $x \, dx = -\frac{1}{2} du$: $I = [\sin^{-1}(x)]_0^1 + \frac{1}{2} \int_1^0 u^{-1/2} \, du$ $I = [\sin^{-1}(x)]_0^1 + [\sqrt{1-x^2}]_0^1$ $I = (\sin^{-1}(1) - \sin^{-1}(0)) + (\sqrt{1-1^2} - \sqrt{1-0^2})$ $I = (\frac{\pi}{2} - 0) + (0 - 1) = \frac{\pi}{2} - 1$

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

Similar Questions

$\int\limits_{ - 1}^{\frac{3}{2}} {|x\sin \pi x|dx} $ equals

$3. \int_0^{\frac{1}{2}} \frac{x \sin^{-1} x}{\sqrt{1-x^2}} dx =$

Let $f(x) = \int\limits_0^x {{e^{ - {t^2}}}dt} $ for all $x > 0$. Then for all $x > 0$:

$\int_{0}^{5} \frac{d x}{x^{2}+2 x+10} = $

Let $f(0)=1, f(0.5)=\frac{5}{4}, f(1)=2, f(1.5)=\frac{13}{4}$ and $f(2)=5$. Using Simpson's rule, $\int_0^2 f(x) dx$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module