evaluate-secxdx-

Question Number 38397 by Fawomath last updated on 25/Jun/18

e v a l u a t e \int s e c x d x

Commented by maxmathsup by imad last updated on 25/Jun/18

l e t I = \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{π}{4} + \frac{x}{2}) ∣ + c

Commented by Cheyboy last updated on 25/Jun/18

\int s e c x d x

\int (s e c x . \frac{s e c x + t a n x}{s e c x + t a n x}) d x

\int (\frac{{s e c}^{2} x + s e c x t a n x}{s e c x + t a n x}) d x

l e t u = s e c x + t a n x

d u = (s e c x t a n x + {s e c}^{2} x) d x

\int \frac{1}{u} d u

\ln ∣ u ∣ + C

\ln ∣ s e c x + t a n x ∣ + C

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Jun/18

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

= \int \frac{c o s x}{{c o s}^{2} x} d x

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

t = s i n x d t = c o s x d x

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

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

= - \frac{1}{2} l n ∣ 1 - t ∣ + \frac{1}{2} l n ∣ 1 + t ∣

= \frac{1}{2} l n ∣ \frac{1 + t}{1 - t} ∣ + c

= \frac{1}{2} l n ∣ \frac{1 + s i n x}{1 - s i n x} ∣ + c

= \frac{1}{2} l n ∣ \frac{{(1 + s i n x)}^{2}}{1 - {s i n}^{2} x} ∣ + c

= \frac{1}{2} l n ∣ \frac{{(1 + s i n x)}^{2}}{{c o s}^{2} x} ∣ + c

= l n ∣ \frac{1}{c o s x} + \frac{s i n x}{c o s x} ∣ + c

= l n ∣ s e c x + t a n x ∣ + c

Answered by malwaan last updated on 25/Jun/18

\int \frac{secx (secx + tanx)}{secx + tanx}

= \int \frac{\sec^{2} x + secx . tanx}{tanx + secx}

= \ln ∣ tanx + secx ∣ + c

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