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

(A) Let $ I = \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx $. Rationalizing the integrand: $ I = \int_{0}^{1} \sqrt{\frac{(1+x)^2}{(1-x)(1+x)}} \, dx = \

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: $ I = \int_{0}^{1} \sqrt{\frac{(1+x)^2}{(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 $ 1-x^2 = t $,then $ -2x \, dx = dt $,so $ x \, dx = -\frac{1}{2} \, dt $. When $ x=0, t=1 $ and when $ x=1, t=0 $. $ \int_{0}^{1} \frac{x}{\sqrt{1-x^2}} \, dx = \int_{1}^{0} \frac{-1/2}{\sqrt{t}} \, dt = \frac{1}{2} \int_{0}^{1} t^{-1/2} \, dt = \frac{1}{2} [2\sqrt{t}]_{0}^{1} = \frac{1}{2} (2) = 1 $. Therefore,$ I = \frac{\pi}{2} + 1 $.

$ \int_{0}^{1} \sqrt{\frac{1+x}{1-x}} \, dx = $

Similar Questions

If $p(x)$ is a cubic polynomial with $p(1)=3, p(0)=2$ and $p(-1)=4$,then $\int_{-1}^1 p(x) dx$ is

$\int_{0}^{\frac{\pi}{2}} \sqrt{\cos \theta} \cdot \sin^{3} \theta d \theta = . . . . . .$

Let $I = \int_{0}^{1} \frac{\sin x}{\sqrt{x}} \, dx$ and $J = \int_{0}^{1} \frac{\cos x}{\sqrt{x}} \, dx$. Then which one of the following is true?

Let $(2^{1-a} + 2^{1+a})$,$f(a)$,$(3^a + 3^{-a})$ be in $A$.$P$. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$ is:

If $f : R \to R$ is a continuous function such that $f(x) = \int\limits_1^x {tf(t)dt}$,then the correct statement is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module