f-x-f-x-dx-ln-sec-x-c-f-x-

Question Number 1597 by Rasheed Ahmad last updated on 25/Aug/15

\int \frac{f (x)}{f^{'} (x)} d x = l n s e c x + c

f (x) = ?

Answered by 123456 last updated on 25/Aug/15

\int \frac{y}{d y / d x} d x = \ln \sec x + c

\frac{d}{d x} [\int \frac{y}{d y / d x} d x] = \frac{d}{d x} [\ln \sec x + c]

\frac{y}{d y / d x} = \frac{\sec x \tan x}{\sec x} = \tan x

y = \frac{d y}{d x} \tan x

\frac{d y}{y} = \frac{d x}{\tan x}

\int \frac{d y}{y} = \int \cot x d x

\ln y = \int \cot x d x = \ln \sin x + g

y = e^{\ln \sin x + g} = e^{\ln \sin x} e^{g}

y = k \sin x, k = e^{g}

- - - - - - - - - - - - - - - - - - - -

verification

y = k \sin x

d y / d x = k \cos x

\int \frac{y}{d y / d x} d x = \int \frac{k \sin x}{k \cos x} d x

= \int \frac{\sin x}{\cos x} d x, u = \cos x, d u = - \sin x d x

= - \int \frac{d u}{u}

= - \ln u + C

= - \ln \cos x + C

= \ln {(\cos x)}^{- 1} + C = \ln \frac{1}{\cos x} + C

= \ln \sec x + C

Commented by 123456 last updated on 25/Aug/15

\frac{d (\sec x)}{d x} = \frac{d}{d x} (\frac{1}{\cos x})

= - \frac{\frac{d (\cos x)}{d x}}{\cos^{2} x}

= \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} = \tan x \sec x

\frac{d}{d x} \ln u = \frac{d u / d x}{u}

\int \cot x d x = \int \frac{\cos x}{\sin x} d x = \int \frac{d u}{u} = \ln u + C = \ln \sin x + C

u = \sin x, d u = \cos x d x

Commented by Rasheed Ahmad last updated on 25/Aug/15

Excellent!