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

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Question

(A) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,so $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Substituting this into the integral: $I =

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

Vedclass · Answer

(A) We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$,so $\cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x$. Substituting this into the integral: $I = \int \frac{\sin^{-1} x - (\frac{\pi}{2} - \sin^{-1} x)}{\frac{\pi}{2}} dx$ $I = \frac{2}{\pi} \int (2 \sin^{-1} x - \frac{\pi}{2}) dx$ $I = \frac{4}{\pi} \int \sin^{-1} x dx - \int dx$ Using integration by parts for $\int \sin^{-1} x dx$,where $u = \sin^{-1} x$ and $dv = dx$: $\int \sin^{-1} x dx = x \sin^{-1} x - \int \frac{x}{\sqrt{1-x^2}} dx = x \sin^{-1} x + \sqrt{1-x^2}$. Substituting this back: $I = \frac{4}{\pi} (x \sin^{-1} x + \sqrt{1-x^2}) - x + c$.

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

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