int sin^{-1}(cos x) , dx = | vedclass

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Question

(C) આપણે જાણીએ છીએ કે $\sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$. તેથી,સંકલન નીચે મુજબ થશે: $\int (\frac{\pi}{2} - x

Vedclass · Accepted Answer

$\frac{\pi x - x^2}{2} + C$

Vedclass · Answer

(C) આપણે જાણીએ છીએ કે $\sin^{-1}(\cos x) = \sin^{-1}(\sin(\frac{\pi}{2} - x)) = \frac{\pi}{2} - x$. તેથી,સંકલન નીચે મુજબ થશે: $\int (\frac{\pi}{2} - x) \, dx$ $= \int \frac{\pi}{2} \, dx - \int x \, dx$ $= \frac{\pi}{2}x - \frac{x^2}{2} + C$ $= \frac{\pi x - x^2}{2} + C$.

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

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