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

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(B) माना $I = \int {{\cos }^{ - 1}}\left( {\frac{1}{x}} \right)\,dx$.चूंकि ${\cos }^{ - 1}\left( {\frac{1}{x}} \right) = {\sec ^{ - 1}}x$,इसलिए $I = \

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) माना $I = \int {{\cos }^{ - 1}}\left( {\frac{1}{x}} \right)\,dx$.चूंकि ${\cos }^{ - 1}\left( {\frac{1}{x}} \right) = {\sec ^{ - 1}}x$,इसलिए $I = \int {{\sec ^{ - 1}}x \cdot 1\,dx}$.खंडशः समाकलन (Integration by parts) का उपयोग करते हुए,$\int {u \cdot v\,dx} = u\int {v\,dx} - \int {\left( {\frac{du}{dx} \cdot \int {v\,dx} } \right)dx}$.माना $u = {\sec ^{ - 1}}x$ और $v = 1$. तब $\frac{du}{dx} = \frac{1}{x\sqrt {{x^2} - 1}}$ और $\int {v\,dx} = x$.$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}$मानक समाकलन $\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}dx} = {\cosh ^{ - 1}}\left( {\frac{x}{a}} \right) + C$ का उपयोग करने पर,हमें प्राप्त होता है:$I = x{\sec ^{ - 1}}x - {\cosh ^{ - 1}}x + C$.

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

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