int {{cos }^{ - 1}}left( {frac{1}{x}} right), | vedclass

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(B) Let $I = \int {{\cos }^{ - 1}}\left( {\frac{1}{x}} \right)\,dx$.Since ${\cos }^{ - 1}\left( {\frac{1}{x}} \right) = {\sec ^{ - 1}}x$,we have $I =

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$x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + C$

Vedclass · Answer

(B) Let $I = \int {{\cos }^{ - 1}}\left( {\frac{1}{x}} \right)\,dx$.Since ${\cos }^{ - 1}\left( {\frac{1}{x}} \right) = {\sec ^{ - 1}}x$,we have $I = \int {{\sec ^{ - 1}}x \cdot 1\,dx}$.Using integration by parts,$\int {u \cdot v\,dx} = u\int {v\,dx} - \int {\left( {\frac{du}{dx} \cdot \int {v\,dx} } \right)dx}$.Let $u = {\sec ^{ - 1}}x$ and $v = 1$. Then $\frac{du}{dx} = \frac{1}{x\sqrt {{x^2} - 1}}$ and $\int {v\,dx} = x$.$I = {\sec ^{ - 1}}x \cdot x - \int {\left( {\frac{1}{x\sqrt {{x^2} - 1}} \cdot x} \right)dx}$$I = x{\sec ^{ - 1}}x - \int {\frac{1}{{\sqrt {{x^2} - 1} }}dx}$Using the standard integral $\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}dx} = {\cosh ^{ - 1}}\left( {\frac{x}{a}} \right) + C$,we get:$I = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + C$.

$\int {{\cos }^{ - 1}}\left( {\frac{1}{x}} \right)\,dx$

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