dx-1-x-2-1-x-2-

Question Number 128775 by bramlexs22 last updated on 10/Jan/21

\int \frac{dx}{{(1 - x)}^{2} \sqrt{1 - x^{2}}} ?

Answered by liberty last updated on 10/Jan/21

let φ = \sqrt{\frac{1 + x}{1 - x}} \Rightarrow x = \frac{φ^{2} - 1}{φ^{2} + 1} and dx = \frac{4 φ}{{(φ^{2} + 1)}^{2}} d φ

Y = \int \frac{{(φ^{2} + 1)}^{3}}{8 φ} \times \frac{4 φ}{{(φ^{2} + 1)}^{2}} d φ

Y = \frac{1}{2} \int (φ^{2} + 1) d φ = \frac{1}{2} (\frac{1}{3} φ^{3} + φ) + C

Y = \frac{1}{6} (\frac{x + 1}{1 - x} + 3) \sqrt{\frac{x + 1}{1 - x}} + C

Y = \frac{2 - x}{3 - 3 x} \sqrt{\frac{x + 1}{1 - x}} + C

Answered by mr W last updated on 10/Jan/21

x = \cos θ

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

= - \int \frac{d θ}{{(1 - \cos θ)}^{2}}

= - \int \frac{d θ}{4 \sin^{4} \frac{θ}{2}}

= - \int \frac{d \frac{θ}{2}}{2 \sin^{4} \frac{θ}{2}}

= \frac{3 \tan^{2} \frac{θ}{2} + 1}{6 \tan^{3} \frac{θ}{2}} + C

= \frac{3 \tan^{2} (\frac{\cos^{- 1} x}{2}) + 1}{6 \tan^{3} (\frac{\cos^{- 1} x}{2})} + C

Answered by mathmax by abdo last updated on 10/Jan/21

\int \frac{dx}{{(1 - x)}^{2} \sqrt{1 - x^{2}}} changement x = sint give

I = \int \frac{cosx}{{(1 - sint)}^{2} cosx} dx =_{\tan (\frac{t}{2}) = u} \int \frac{2 du}{(1 + u^{2}) {(1 - \frac{2 u}{1 + u^{2}})}^{2}}

= \int \frac{2 du}{(u^{2} + 1) {(\frac{u^{2} + 1 - 2 u}{u^{2} + 1})}^{2}} = 2 \int \frac{u^{2} + 1}{{(u^{2} - 2 u + 1)}^{2}} du

= 2 \int \frac{u^{2} - 2 u + 1 + 2 u}{{(u^{2} - 2 u + 1)}^{2}} du = 2 \int \frac{du}{(u^{2} - 2 u + 1)} + 2 \int \frac{2 u - 2 + 2}{{(u^{2} - 2 u + 1)}^{2}} du

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

= - \frac{2}{u - 1} - \frac{2}{u^{2} - 2 u + 1} + 4 \times \frac{1}{- 4 + 1} {(u - 1)}^{- 4 + 1} + C

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

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

= \frac{10}{3 (1 - \tan (\frac{arcsinx}{2}))} - \frac{2}{{(1 - \tan (\frac{arcsinx}{2}))}^{2}} + C