1-sec-x-dx-

Question Number 133360 by liberty last updated on 21/Feb/21

\int \sqrt{1 + \sec x} dx ?

Commented by som(math1967) last updated on 21/Feb/21

\int \sqrt{\frac{1 + c o s x}{c o s x}} d x

\sqrt{2} \int \frac{c o s \frac{x}{2}}{\sqrt{1 - 2 {s i n}^{2} \frac{x}{2}}} d x

= \frac{\sqrt{2}}{\sqrt{2}} \int \frac{c o s \frac{x}{2}}{\sqrt{{(\frac{1}{\sqrt{2}})}^{2} - {s i n}^{2} \frac{x}{2}}} d x

= 2 \int \frac{c o s \frac{x}{2}}{2 \sqrt{{(\frac{1}{\sqrt{2}})}^{2} - {s i n}^{2} \frac{x}{2}}} d x

= 2 \int \frac{d (s i n \frac{x}{2})}{\sqrt{{(\frac{1}{\sqrt{2}})}^{2} - {s i n}^{2} \frac{x}{2}}}

= {2 \sin}^{- 1} \frac{s i n \frac{x}{2}}{\frac{1}{\sqrt{2}}} + c

= {2 \sin}^{- 1} \sqrt{2} s i n \frac{x}{2} + c

Commented by liberty last updated on 22/Feb/21

in other way

I = \int \frac{\sqrt{1 + \cos x}}{\sqrt{\cos x}} dx

let \cos x = z^{2} \to dx = - \frac{2 z}{\sin x} dz

dx = - \frac{2 z}{\sqrt{1 - \cos^{2} x}} dz = - \frac{2 z}{\sqrt{1 - z^{4}}} dz

I = \int \frac{\sqrt{1 + z^{2}}}{z} (- \frac{2 z}{\sqrt{(1 + z^{2}) (1 - z^{2})}} dz)

I = - 2 \int \frac{dz}{\sqrt{1 - z^{2}}} = - {2 \sin}^{- 1} (z) + c

I = - {2 \sin}^{- 1} (\sqrt{\cos x}) + c

Answered by liberty last updated on 21/Feb/21

Commented by liberty last updated on 21/Feb/21

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