int frac{1}{(cos^{-1} x) sqrt{1 - x^2}} dx = | vedclass

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(B) Let $t = \cos^{-1} x$. Then,differentiating both sides with respect to $x$,we get $dt = -\frac{1}{\sqrt{1 - x^2}} dx$,which implies $\frac{dx}{\sq

Vedclass · Accepted Answer

$-\log(\cos^{-1} x) + c$

Vedclass · Answer

(B) Let $t = \cos^{-1} x$. Then,differentiating both sides with respect to $x$,we get $dt = -\frac{1}{\sqrt{1 - x^2}} dx$,which implies $\frac{dx}{\sqrt{1 - x^2}} = -dt$. Substituting these into the integral: $\int \frac{1}{(\cos^{-1} x) \sqrt{1 - x^2}} dx = \int \frac{1}{t} (-dt) = -\int \frac{1}{t} dt$. Integrating,we get $-\log|t| + c$. Substituting $t = \cos^{-1} x$ back,the result is $-\log(\cos^{-1} x) + c$.

$\int \frac{1}{(\cos^{-1} x) \sqrt{1 - x^2}} dx = $

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