∫secxdx ?


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Question Number 63976 by Scientist0000001 last updated on 11/Jul/19

$\int s e c x d x ?$
Commented by Prithwish sen last updated on 12/Jul/19

$\int \frac{\sec^{2} \frac{x}{2}}{1 - \tan^{2} \frac{x}{2}} dx putting \tan \frac{x}{2} = t$ $\sec^{2} \frac{x}{2} dx = 2 dt ∴ 2 \int \frac{dt}{1 - t^{2}} = \int [\frac{1}{1 - t} + \frac{1}{1 + t}] dt$ $= \ln ∣ \frac{1 + t}{1 - t} ∣ + C$
Commented by hknkrc46 last updated on 11/Jul/19
$∫secxdx=∫(dx/(cosx))=∫((cosx)/(cos^2 x))dx=∫((cosx)/(1−sin^2 x))dx sinx=u⇒cosxdx=du ∫(du/(1−u^2 ))=(1/2)∫(((1−u)+(1+u))/((1−u)(1+u)))du =(1/2)[∫((1−u)/((1−u)(1+u)))du+∫((1+u)/((1−u)(1+u)))du] =(1/2)[∫(du/(1+u))+∫(du/(1−u))]=(1/2)[ln (1+u)−ln (1−u)]+c =(1/2)ln (((1+u)/(1−u)))+c=(1/2)ln (((1+sinx)/(1−sinx)))+c ★ =(1/2)ln ((((1+sinx)^2 )/((1+sinx)(1−sinx))))+c =(1/2)ln ((((1+sinx)^2 )/(1−sin^2 x)))+c =(1/2)ln ((((1+sinx)^2 )/(cos^2 x)))+c=ln (((1+sinx)/(cosx)))+c =ln ((1/(cosx))+((sinx)/(cosx)))+c=ln (secx+tanx)+c ★★ =ln (secx+tanx)+ln e^c =ln (e^c (secx+tanx)) =ln (c_1 (secx+tanx)) ★★★ {c , c_1 ∈R }$
$\int s e c x d x = \int \frac{d x}{c o s x} = \int \frac{c o s x}{{c o s}^{2} x} d x = \int \frac{c o s x}{1 - {s i n}^{2} x} d x$ $s i n x = u \Rightarrow c o s x d x = d u$ $\int \frac{d u}{1 - u^{2}} = \frac{1}{2} \int \frac{(1 - u) + (1 + u)}{(1 - u) (1 + u)} d u$ $= \frac{1}{2} [\int \frac{1 - u}{(1 - u) (1 + u)} d u + \int \frac{1 + u}{(1 - u) (1 + u)} d u]$ $= \frac{1}{2} [\int \frac{d u}{1 + u} + \int \frac{d u}{1 - u}] = \frac{1}{2} [\ln (1 + u) - \ln (1 - u)] + c$ $= \frac{1}{2} \ln (\frac{1 + u}{1 - u}) + c = \frac{1}{2} \ln (\frac{1 + s i n x}{1 - s i n x}) + c ★$ $= \frac{1}{2} \ln (\frac{{(1 + s i n x)}^{2}}{(1 + s i n x) (1 - s i n x)}) + c$ $= \frac{1}{2} \ln (\frac{{(1 + s i n x)}^{2}}{1 - {s i n}^{2} x}) + c$ $= \frac{1}{2} \ln (\frac{{(1 + s i n x)}^{2}}{{c o s}^{2} x}) + c = \ln (\frac{1 + s i n x}{c o s x}) + c$ $= \ln (\frac{1}{c o s x} + \frac{s i n x}{c o s x}) + c = \ln (s e c x + t a n x) + c ★ ★$ $= \ln (s e c x + t a n x) + \ln e^{c} = \ln (e^{c} (s e c x + t a n x))$ $= \ln (c_{1} (s e c x + t a n x)) ★ ★ ★$ ${c, c_{1} \in R}$
Commented by mathmax by abdo last updated on 11/Jul/19

$l e t I = \int \frac{d x}{c o s x} c h a n g e m e n t t a n (\frac{x}{2}) = t g i v e$ $I = \int \frac{2 d t}{(1 + t^{2}) \frac{1 - t^{2}}{1 + t^{2}}} = \int \frac{2 d t}{1 - t^{2}} = \int (\frac{1}{1 - t} + \frac{1}{1 + t}) d t$ $l n ∣ \frac{1 + t}{1 - t} ∣ + c = l n ∣ \frac{1 + t a n (\frac{x}{2})}{1 - t a n (\frac{x}{2})} ∣ + c = l n ∣ t a n (\frac{π}{4} + \frac{x}{2}) ∣ + c .$
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