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Question Number 40717 by ajeetyadav4370 last updated on 26/Jul/18

\int \sqrt{t a n x / s i n x . c o s x d x}

Commented by math khazana by abdo last updated on 26/Jul/18

l e t I = \int \sqrt{\frac{t a n x}{s i n x c o s x}} d x

I = \int \sqrt{\frac{s i n x}{s i n x {c o s}^{2} x}} d x

= \int \frac{d x}{c o s x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e

I = \int \frac{1}{\frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d t}{1 + t^{2}} = \int \frac{2 d t}{1 - t^{2}}

= \int {\frac{1}{1 + t} + \frac{1}{1 - t}} d t = l n ∣ \frac{1 + t}{1 - t} ∣ + c

= l n ∣ \frac{1 + t a n (\frac{x}{2})}{1 - t a n (\frac{x}{2})} ∣ + c = l n ∣ t a n (\frac{x}{2} + \frac{π}{4}) ∣ + c .

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Jul/18

\int \sqrt{\frac{t a n x}{s i n x c o s x}} d x

\int s e c x d x

\int \frac{d x}{c o s x}

\int \frac{1 + {t a n}^{2} \frac{x}{2}}{1 - {t a n}^{2} \frac{x}{2}} d x l e t t = t a n \frac{x}{2} d t = {s e c}^{2} \frac{x}{2} . \frac{1}{2} . d x

\int \frac{1 + t^{2}}{1 - t^{2}} \times \frac{2 d t}{1 + t^{2}}

2 \int \frac{d t}{1 - t^{2}}

\int \frac{1 + t + 1 - t}{(1 + t) (1 - t)} d t

\int \frac{d t}{1 - t} + \int \frac{d t}{1 + t}

l n (1 + t) - l n (1 - t) + c

l n (\frac{1 + t}{1 - t}) + c

l n (\frac{1 + t a n \frac{x}{2}}{1 - t a n \frac{x}{2}}) + c

l n (\frac{1 + 2 t a n \frac{x}{2} + {t a n}^{2} \frac{x}{2}}{1 - {t a n}^{2} \frac{x}{2}}) + c

l n (\frac{1 + \frac{2 t a n \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}}{\frac{1 - {t a n}^{2} \frac{x}{2}}{1 + {t a n}^{2} \frac{x}{2}}}) = l n (\frac{1 + s i n x}{c o s x}) = l n (s e c x + t a n x)

Commented by $@ty@m last updated on 26/Jul/18

\int \sec x d x

\int \sec x . \frac{(\sec x + \tan x)}{(\sec x + \tan x)} d x

= \ln (\sec x + \tan x) + C

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