int {frac{{{{sin }^{ - 1}}x - {{cos }^{ - 1}} | vedclass

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

Question

(A) આપણે જાણીએ છીએ કે $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,તેથી $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. આ કિંમત સંકલનમાં મૂકતા: $I = \int

Vedclass · Accepted Answer

$\frac{4}{\pi }\left( {x{{\sin }^{ - 1}}x + \sqrt {1 - {x^2}} } \right) - x + c$

Vedclass · Answer

(A) આપણે જાણીએ છીએ કે $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,તેથી $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. આ કિંમત સંકલનમાં મૂકતા: $I = \int \frac{\sin^{-1} x - (\frac{\pi}{2} - \sin^{-1} x)}{\frac{\pi}{2}} dx$ $I = \frac{2}{\pi} \int (2 \sin^{-1} x - \frac{\pi}{2}) dx$ $I = \frac{4}{\pi} \int \sin^{-1} x dx - \int dx$ $\int \sin^{-1} x dx$ માટે ખંડશઃ સંકલનનો ઉપયોગ કરતા,જ્યાં $u = \sin^{-1} x$ અને $dv = dx$: $\int \sin^{-1} x dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx = x \sin^{-1} x + \sqrt{1-x^2}$. આ કિંમત પાછી મૂકતા: $I = \frac{4}{\pi} (x \sin^{-1} x + \sqrt{1-x^2}) - x + c$.

List-$I$	List-$II$
$1. \int \frac{\sin^2 x}{\cos^4 x} dx$	$A. \frac{\tan^2 x}{2} + \ln\|\cos x\| + c$
$2. \int \frac{\sin^4 x}{\cos^2 x} dx$	$B. \cos x + \sec x + c$
$3. \int \frac{\sin^3 x}{\cos^2 x} dx$	$C. \frac{\tan^3 x}{3} + c$
$4. \int \frac{\sin^3 x}{\cos^3 x} dx$	$D. \tan x + \frac{\sin 2x}{4} - \frac{3x}{2} + c$
	$E. \cos x - \sec x + c$

$\int {\frac{{{{\sin }^{ - 1}}x - {{\cos }^{ - 1}}x}}{{{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x}}} dx = $

Similar Questions

$\int (\sin^{-1} x + \cos^{-1} x) \, dx = $

$\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x\left(1+\cos ^2 2 x\right)} d x=$

$\int \frac{\cos 2 x \cdot \sin 4 x}{\cos ^4 x(1+\cos ^2 2 x)} d x=$

$y=\int \cos \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} d x$ એ કોનું સમીકરણ છે?

Vedclass Test Series

Exam Paper Generator

Online Exam Module