If int e^xleft(frac{x sin ^{-1} x}{sqrt{1-x^2 | vedclass

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

Question

(C) We know that $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.Let $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.Then $f'(x) = \frac{d}{dx} \left( \frac{x

Vedclass · Accepted Answer

$\frac{\pi}{6} \sqrt{\frac{e}{3}}$

Vedclass · Answer

(C) We know that $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.Let $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.Then $f'(x) = \frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1-x^2}} \right) = \frac{\sqrt{1-x^2} \cdot \frac{d}{dx}(x \sin^{-1} x) - x \sin^{-1} x \cdot \frac{d}{dx}(\sqrt{1-x^2})}{1-x^2}$$= \frac{\sqrt{1-x^2} \left( \sin^{-1} x + \frac{x}{\sqrt{1-x^2}} \right) - x \sin^{-1} x \left( \frac{-x}{\sqrt{1-x^2}} \right)}{1-x^2}$$= \frac{\sqrt{1-x^2} \sin^{-1} x + x + \frac{x^2 \sin^{-1} x}{\sqrt{1-x^2}}}{1-x^2} = \frac{(1-x^2) \sin^{-1} x + x \sqrt{1-x^2} + x^2 \sin^{-1} x}{(1-x^2)^{3/2}} = \frac{\sin^{-1} x}{(1-x^2)^{3/2}} + \frac{x}{1-x^2}$.Thus,the integral is $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C = e^x \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + C$.So,$g(x) = \frac{x e^x \sin^{-1} x}{\sqrt{1-x^2}}$.Evaluating at $x = \frac{1}{2}$: $g\left(\frac{1}{2}\right) = \frac{\frac{1}{2} e^{1/2} \sin^{-1}(1/2)}{\sqrt{1-(1/2)^2}} = \frac{\frac{1}{2} \sqrt{e} \cdot \frac{\pi}{6}}{\sqrt{3/4}} = \frac{\frac{\pi \sqrt{e}}{12}}{\frac{\sqrt{3}}{2}} = \frac{\pi \sqrt{e}}{6 \sqrt{3}} = \frac{\pi}{6} \sqrt{\frac{e}{3}}$.

If $\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) d x=g(x)+C$ where $C$ is the constant of integration,then $g \left(\frac{1}{2}\right)$ equals :

Similar Questions

$\int \frac{e^{\tan ^{-1} x}}{1+x^2}\left[\left(\sec ^{-1} \sqrt{1+x^2}\right)^2+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right] d x$,where $x>0$ is

$\int \left(1+x-\frac{1}{x}\right) e^{x+\frac{1}{x}} \,d x$ is equal to

$\int {{e^x}\left[ {f(x) + f'(x)} \right]\,dx} $ is equal to

$\int \left( \frac{1-\log x}{1+(\log x)^2} \right)^2 dx = $

$\int \frac{x+100}{(x+101)^2} e^x \, dx = $ . . . . . . $+ C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module