sec-x-1-2sec-x-cosec-x-cot-x-cosec-x-cot-x-dx-

Question Number 160792 by cortano last updated on 06/Dec/21

\int \frac{\sec x}{\sqrt{1 + 2 \sec x}} \sqrt{\frac{cosec x - \cot x}{cosec x + \cot x}} dx = ?

Answered by chhaythean last updated on 06/Dec/21

= \int \frac{\frac{1}{cosx}}{\sqrt{\frac{cosx + 2}{cosx}}} \times \sqrt{\frac{\frac{1 - cosx}{sinx}}{\frac{1 + cosx}{sinx}}} dx

= \int \frac{\frac{1}{cosx}}{\sqrt{\frac{cosx + 2}{cosx}}} \times \sqrt{\frac{1 - cosx}{1 + cosx}} dx

= \int \frac{sinx}{\sqrt{\cos^{2} x + 2 cosx}} \times \frac{1}{1 + cosx} dx

= \int \frac{sinx}{\sqrt{{(cosx + 1)}^{2} - 1} (1 + cosx)} dx

let u = 1 + cosx \Rightarrow du = - sinxdx

= - \int \frac{du}{\sqrt{u^{2} - 1} \times u}

let u = secy \Rightarrow du = secytanydy

= - \int \frac{secytany}{tanysecy} dy = - y + c

= - arcsec (u) + c

= - arcsec (1 + cosx) + c

So \begin{matrix} \int \frac{secx}{\sqrt{1 + 2 secx}} \times \sqrt{\frac{cosecx - cotx}{cosecx + cotx}} dx = - arcsec (1 + cosx) + c \end{matrix}

Answered by MJS_new last updated on 06/Dec/21

let c = \cos x \to d x = - \frac{d c}{\sqrt{1 - c^{2}}}

now we have

- \int \frac{d c}{(c + 1) \sqrt{c} \sqrt{c + 2}} =

[t = \frac{\sqrt{c + 2}}{\sqrt{c}} \to d c = - \sqrt{c^{3}} \sqrt{c + 2} d t]

= 2 \int \frac{d t}{t^{2} + 1} = 2 \arctan t = 2 \arctan \frac{\sqrt{c + 2}}{\sqrt{c}} =

= 2 \arctan \frac{\sqrt{2 + \cos x}}{\sqrt{\cos x}} + C

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