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

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

Question

(C) हम जानते हैं कि $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.माना $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.तब $f'(x) = \frac{d}{dx} \left( \frac

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) हम जानते हैं कि $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$.माना $f(x) = \frac{x \sin^{-1} x}{\sqrt{1-x^2}}$.तब $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}$.अतः,समाकलन $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C = e^x \frac{x \sin^{-1} x}{\sqrt{1-x^2}} + C$ है।इसलिए,$g(x) = \frac{x e^x \sin^{-1} x}{\sqrt{1-x^2}}$.$x = \frac{1}{2}$ पर मान रखने पर: $g\left(\frac{1}{2}\right) = \frac{\frac{1}{2} e^{1/2} \sin^{-1}(1/2)}{\sqrt{1-(1/2)^2}} = \frac{\frac{1}{2} \sqrt{e} \cdot \frac{\pi}{6}}{\sqrt{3/4}} = \frac{\frac{\pi \sqrt{e}}{12}}{\frac{\sqrt{3}}{2}} = \frac{\pi \sqrt{e}}{6 \sqrt{3}} = \frac{\pi}{6} \sqrt{\frac{e}{3}}$.

यदि $\int e^x\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\frac{\sin ^{-1} x}{\left(1-x^2\right)^{3 / 2}}+\frac{x}{1-x^2}\right) d x=g(x)+C$ जहाँ $C$ समाकलन स्थिरांक है,तो $g \left(\frac{1}{2}\right)$ का मान ज्ञात कीजिए :

Similar Questions

$\int e^x \left( \frac{1+\sin x}{1+\cos x} \right) dx =$

$\int {{e^{2x}}( - \sin x + 2\cos x)\,dx} = $

$\int e^{x / 2}\left(\frac{2+\sin x}{1+\cos x}\right) d x=$

$ \int \frac{e^{x}\left(x^{2} \tan ^{-1} x+\tan ^{-1} x+1\right)}{x^{2}+1} d x $ का मान ज्ञात कीजिए।

$\int_1^{e} \frac{e^x}{x}(1+x \log x) d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module