dx-1-sinx-2-

Question Number 103825 by mohammad17 last updated on 17/Jul/20

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

Answered by ~blr237~ last updated on 17/Jul/20

\int \frac{1 + 2 s i n x + (1 - {c o s}^{2} x)}{{c o s}^{4} x} d x

\int [2 (1 + {t a n}^{2} x) + 2 (1 + {t a n}^{2} x) {t a n}^{2} x + \frac{2 s i n x}{{(c o s x)}^{4}} - \frac{1}{{c o s}^{2} x}) d x

Commented by mohammad17 last updated on 17/Jul/20

h o w t h i s s i r

Answered by mathmax by abdo last updated on 17/Jul/20

let f (a) = \int \frac{dx}{a - sinx} we have f^{^{'}} (a) = - \int \frac{dx}{{(a - sinx)}^{2}} \Rightarrow

\int \frac{dx}{{(a - sinx)}^{2}} = - f^{^{'}} (a) and \int \frac{dx}{{(1 - sinx)}^{2}} = - f^{^{'}} (1) let explicite f (a)

changement \tan (\frac{x}{2}) = t give f (a) = \int \frac{2 dt}{(1 + t^{2}) (a - \frac{2 t}{1 + t^{2}})}

= \int \frac{2 dt}{a + {at}^{2} - 2 t} = \int \frac{2 dt}{{at}^{2} - 2 t + a}

Δ^{^{'}} = 1 - a^{2} let take a > 1 \Rightarrow Δ^{^{'}} < 0 \Rightarrow f (a) = \frac{2}{a} \int \frac{dt}{t^{2} - \frac{2 t}{a} + 1}

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

=_{t - \frac{1}{a} = \frac{\sqrt{a^{2} - 1}}{a} u} \frac{2}{a} \times \frac{a^{2}}{a^{2} - 1} \int \frac{1}{1 + u^{2}} \times \frac{\sqrt{a^{2} - 1}}{a} du

= \frac{2}{\sqrt{a^{2} - 1}} \int \frac{du}{1 + u^{2}} = \frac{2}{\sqrt{a^{2} - 1}} \arctan (\frac{at - 1}{\sqrt{a^{2} - 1}}) = \frac{2}{\sqrt{a^{2} - 1}} \arctan (\frac{atan (\frac{x}{2}) - 1}{\sqrt{a^{2} - 1}}) + c

rest to calculate f^{^{'}} (a) and \lim_{a \to 1} f^{^{'}} (a) \dots be continued \dots

Answered by Dwaipayan Shikari last updated on 17/Jul/20

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

\int s e c x (s e c x + t a n x) s e c x (s e c x + t a n x) d x

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

\int \frac{t}{\sqrt{2 t^{2} - t^{4}}} d t {\frac{1 + s i n x}{c o s x} = t \Rightarrow 1 + {s i n}^{2} x + 2 s i n x = t^{2} (1 - {s i n}^{2} x)

\int \frac{1}{\sqrt{2 - t^{2}}} d t {\Rightarrow (1 + t^{2}) {s i n}^{2} x + 2 s i n x + 1 - t^{2} = 0

\int \frac{\sqrt{2} c o s θ}{\sqrt{2 - 2 {s i n}^{2} θ}} d θ {t = \sqrt{2} s i n θ \sqrt{2} c o s θ = \frac{d t}{d θ}

\int d θ {s i n x = \frac{- 2 \pm \sqrt{4 - 4 (1 - t^{4})}}{2} = t^{2} - 1

= θ + C

{c o s x = \sqrt{1 - {(t^{2} - 1)}^{2}}

= {s i n}^{- 1} \frac{t}{\sqrt{2}} + C

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

Commented by mohammad17 last updated on 17/Jul/20

t h a n k y o u s i r

Commented by Dwaipayan Shikari last updated on 17/Jul/20

K i n d l y c h e c k i t

Commented by Dwaipayan Shikari last updated on 17/Jul/20

Commented by Dwaipayan Shikari last updated on 17/Jul/20

A l t e r n a t e s o l u t i o n

Answered by mathmax by abdo last updated on 17/Jul/20

let try another way

I = \int \frac{dx}{{(1 - sinx)}^{2}} \Rightarrow I = \int \frac{dx}{{(1 - \cos (\frac{π}{2} - x))}^{2}} = \int \frac{dx}{{({2 \sin}^{2} (\frac{π}{4} - \frac{x}{2}))}^{4}}

= \frac{1}{16} \int \frac{dx}{\sin^{4} (\frac{π}{4} - \frac{x}{2})} =_{\frac{π}{4} - \frac{x}{2} = t} \frac{1}{16} \int \frac{- 2 dt}{\sin^{4} t} = - \frac{1}{8} \int \frac{dt}{\sin^{4} t}

changement \tan (\frac{t}{2}) = u give \int \frac{dt}{\sin^{4} t} = \int \frac{2 du}{(1 + u^{2}) {(\frac{2 u}{1 + u^{2}})}^{4}}

= \int \frac{2 {(1 + u)}^{3}}{{16 u}^{4}} du = \frac{1}{8} \int \frac{u^{3} + {3 u}^{2} + 3 u + 1}{u^{4}} du

= \frac{1}{8} \int (\frac{1}{u} + \frac{3}{u^{2}} + \frac{3}{u^{3}} + \frac{1}{u^{4}}) du

= \frac{1}{8} {\ln ∣ u ∣ - \frac{3}{u} + 3 (- \frac{1}{{2 u}^{2}}) - \frac{1}{{3 u}^{3}}} + C

\Rightarrow I = - \frac{1}{64} {\ln ∣ \tan (\frac{t}{2}) ∣ - \frac{3}{\tan (\frac{t}{2})} - \frac{3}{{2 \tan}^{2} (\frac{t}{2})} - \frac{1}{{3 \tan}^{3} (\frac{t}{2})}} + C

= - \frac{1}{64} {\ln ∣ \tan (\frac{π}{8} - \frac{x}{4}) ∣ - \frac{3}{\tan (\frac{π}{8} - \frac{x}{4})} - \frac{3}{{2 \tan}^{2} (\frac{π}{8} - \frac{x}{4})} - \frac{1}{{3 \tan}^{3} (\frac{π}{8} - \frac{x}{4})}} + C

Commented by mathmax by abdo last updated on 17/Jul/20

error of typo I = \int \frac{dx}{{({2 \sin}^{2} (\frac{π}{4} - \frac{x}{2}))}^{2}} = \dots

Commented by mathmax by abdo last updated on 17/Jul/20

sorry change 64 by 32 \dots

Commented by mohammad17 last updated on 17/Jul/20

o k s i r t h a n k y o u

Answered by PRITHWISH SEN 2 last updated on 17/Jul/20

\int \frac{dx}{{(1 - \frac{2 \tan \frac{x}{2}}{1 + \tan^{2} \frac{x}{2}})}^{2}} = \int \frac{\sec^{2} \frac{x}{2}}{{(1 - \tan \frac{x}{2})}^{4}} (1 + \tan^{2} \frac{x}{2}) dx

\tan \frac{x}{2} = t \Rightarrow \sec^{2} \frac{x}{2} dx = 2 dt

= 2 \int \frac{(1 + t^{2}) dt}{{(1 - t)}^{4}} = \int \frac{2 dt}{{(1 - t)}^{2}} - \int \frac{4 (1 - t) dt}{{(1 - t)}^{4}} + 4 \int \frac{dt}{{(1 - t)}^{4}}

= \frac{2}{(1 - t)} + \frac{2}{{(1 - t)}^{2}} - \frac{4}{3 {(1 - t)}^{3}} + C

= \frac{2}{(1 - \tan \frac{x}{2})} + \frac{2}{{(1 - \tan \frac{x}{2})}^{2}} - \frac{4}{3 {(1 - \tan \frac{x}{2})}^{3}} + C

please check

Commented by mohammad17 last updated on 17/Jul/20

t h a n y o u s i r

Answered by Ar Brandon last updated on 19/Jul/20

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

= \int \frac{1 + 2 sinx + \sin^{2} x}{{(1 - \sin^{2} x)}^{2}} dx = \int \frac{\cos^{2} x + 2 sinx + {2 \sin}^{2} x}{\cos^{4} x} dx

= \int \sec^{2} xdx + 2 \int \frac{sinx}{\cos^{4} x} dx + 2 \int \tan^{2} {xsec}^{2} xdx

= tanx + \frac{{2 \sec}^{3} x}{3} + \frac{{2 \tan}^{3} x}{3} + C

Answered by Her_Majesty last updated on 07/Aug/20

\int \frac{d x}{{(1 - \sin x)}^{2}}

simply {Weyerstrass}^{'} Substitution

t = \tan \frac{x}{2}

leads to

2 \int \frac{t^{2} + 1}{{(t - 1)}^{4}} d t = - \frac{2 (3 t^{2} - 3 t + 2)}{3 {(t - 1)}^{3}} = \dots