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

Question Number 119920 by bramlexs22 last updated on 28/Oct/20

\int \sqrt{\frac{1 - x}{1 + x}} d x = ?, x ϵ (- 1, 1)

Answered by mathmax by abdo last updated on 28/Oct/20

I = \int \sqrt{\frac{1 - x}{1 + x}} dx we do the changement x = cost \Rightarrow

I = \int \sqrt{\frac{{2 \sin}^{2} (\frac{t}{2})}{{2 \cos}^{2} (\frac{t}{2})}} sint dt

= \int \frac{\sin (\frac{t}{2})}{\cos (\frac{t}{2})} \times 2 \sin (\frac{t}{2}) \cos (\frac{t}{2}) dt

= 2 \int \sin^{2} (\frac{t}{2}) dt = 2 \int \frac{1 - \cos (t)}{2} dt = \int (1 - cost) dt

= t - sint + c = arcosx - \sqrt{1 - x^{2}} + c = arcsinx - \sqrt{1 - x^{2}} + C

Answered by 1549442205PVT last updated on 28/Oct/20

\int \sqrt{\frac{1 - x}{1 + x}} d x = ?, x ϵ (- 1, 1)

Put x = \cos φ \Rightarrow dx = - \sin φ d φ

F = - \int \sqrt{\frac{1 - \cos φ}{1 + \cos φ}} . \sin φ d φ

= - \int \sqrt{\frac{{2 \sin}^{2} \frac{φ}{2}}{{2 \cos}^{2} \frac{φ}{2}}} . \sin φ d φ

= - \int \tan \frac{φ}{2} . 2 \sin \frac{φ}{2} \cos \frac{φ}{2} d φ

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

= - φ + \sin φ = - \cos^{- 1} (x) + \sqrt{1 - x^{2}} + C

Answered by bobhans last updated on 28/Oct/20

\int \sqrt{\frac{1 - x}{1 + x}} d x = ?, x \in (- 1, 1)

K = \int \sqrt{\frac{1 - x}{1 + x}} d x [l e t \sqrt{1 + x} = 1 + t]

1 + x = 1 + 2 t + t^{2} \Rightarrow d x = (2 + 2 t) d t

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

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

b y s u b s t i t u t i n g 1 + t = \sqrt{2} \sin w

K = 4 \int \cos^{2} w d w = 4 \int (\frac{1}{2} + \frac{1}{2} \cos 2 w) d w

K = 2 w + \sin 2 w + c

K = 2 arc \sin (\frac{1 + t}{\sqrt{2}}) + 2 (\frac{1 + t}{\sqrt{2}}) (\frac{\sqrt{1 - 2 t - t^{2}}}{\sqrt{2}}) + c

K = 2 arc \sin (\sqrt{\frac{1 + x}{2}}) + \sqrt{1 - x^{2}} + c

Answered by Ar Brandon last updated on 28/Oct/20

I = \int \sqrt{\frac{1 - x}{1 + x}} dx = \int \frac{1 - x}{\sqrt{1 - x^{2}}} dx

= \int \frac{dx}{\sqrt{1 - x^{2}}} - \int \frac{x}{\sqrt{1 - x^{2}}} dx

= Arcsin (x) + \sqrt{1 - x^{2}} + C

Answered by Dwaipayan Shikari last updated on 28/Oct/20

\int \frac{1 - x}{\sqrt{1 - x^{2}}} d x

= \int \frac{1}{\sqrt{1 - x^{2}}} + \frac{1}{2} \int \frac{- 2 x}{\sqrt{1 - x^{2}}} d x

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