sin-1-x-a-x-dx-a-gt-0-

Question Number 85909 by jagoll last updated on 26/Mar/20

\int \sin^{- 1} (\sqrt{\frac{x}{a + x}}) dx, a > 0

Commented by john santu last updated on 26/Mar/20

l e t \sqrt{\frac{x}{a + x}} = n \Rightarrow x = \frac{{a n}^{2}}{1 - n^{2}}

I = \int \sin^{- 1} (n) \times \frac{2 a n}{{(1 - n^{2})}^{2}} d n

[h = \sin^{- 1} (n) \Rightarrow n = \sin h]

I = 2 a \int h \tan h \sec^{2} h d h

I = a (h \tan^{2} h - \int (\sec^{2} h - 1) d h)

I = a (h \tan^{2} h - \tan h + h) + c

∴ I = x \sin^{- 1} (\sqrt{\frac{x}{a + x}}) - \sqrt{a x} + a \sin^{- 1} (\sqrt{\frac{x}{a + x}}) + c

Commented by jagoll last updated on 26/Mar/20

thank you all

Answered by TANMAY PANACEA. last updated on 26/Mar/20

x = {a t a n}^{2} α \to \frac{d x}{d α} = a \times 2 t a n α \times {s e c}^{2} α

\sqrt{\frac{{a t a n}^{2} α}{a + {a t a n}^{2} α}} = \sqrt{\frac{{s i n}^{2} α}{{c o s}^{2} α \times {s e c}^{2} α}} = s i n α

I = \int α . 2 a t a n α . {s e c}^{2} α d α

\frac{I}{2 a} = α \int t a n α {s e c}^{2} α d α - \int [\frac{d α}{d α} \int t a n α {s e c}^{2} α d α] d α

= α . \frac{{t a n}^{2} α}{2} - \frac{1}{2} \int ({s e c}^{2} α - 1) d α

= \frac{α {t a n}^{2} α}{2} - \frac{t a n α}{2} + \frac{α}{2}

n o w α = {t a n}^{- 1} \sqrt{\frac{x}{a}}

I = (\frac{{t a n}^{- 1} \sqrt{\frac{x}{a}} \times \frac{x}{a}}{2} - \frac{\sqrt{\frac{x}{a}}}{2} + \frac{{t a n}^{- 1} \sqrt{\frac{x}{a}}}{2}) \times 2 a

I = {x t a n}^{- 1} \sqrt{\frac{x}{a}} - a \sqrt{\frac{x}{a}} + {a t a n}^{- 1} \sqrt{\frac{x}{a}}

= (x + a) {t a n}^{- 1} \sqrt{\frac{x}{a}} - \sqrt{a x}

p l s c h e c k

Answered by Kunal12588 last updated on 26/Mar/20

I = {x s i n}^{- 1} \sqrt{\frac{x}{a + x}} - \int x \times \frac{1}{\sqrt{1 - \frac{x}{a + x}}} \times \frac{1}{2 \sqrt{\frac{x}{a + x}}} \times \frac{a + x - x}{{(a + x)}^{2}} d x

= {x s i n}^{- 1} \sqrt{\frac{x}{a + x}} - \int x \times \frac{\sqrt{a + x}}{\sqrt{a}} \times \frac{\sqrt{a + x}}{2 \sqrt{x}} \times \frac{a}{{(a + x)}^{2}} d x

= {x s i n}^{- 1} \sqrt{\frac{x}{a + x}} - \frac{1}{2} \sqrt{a} \int \sqrt{x} \times \frac{1}{a + x} d x

I_{1} = \int \frac{\sqrt{x}}{a + x} d x

\sqrt{x} = t

d x = 2 \sqrt{x} d t

I_{1} = 2 \int \frac{x}{a + x} d t = 2 \int \frac{t^{2}}{a + t^{2}} d t

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

= 2 t - 2 a \times \frac{1}{\sqrt{a}} {t a n}^{- 1} \frac{t}{\sqrt{a}} + c

= 2 \sqrt{x} - 2 \sqrt{a} {t a n}^{- 1} \frac{\sqrt{x}}{\sqrt{a}} + c

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

I = {x s i n}^{- 1} \sqrt{\frac{x}{a + x}} - \sqrt{a x} + a {t a n}^{- 1} \sqrt{\frac{x}{a}} + C

Commented by Kunal12588 last updated on 26/Mar/20

{s i n}^{- 1} \sqrt{\frac{x}{a + x}} = {t a n}^{- 1} \sqrt{\frac{x}{a}}

s o T a n m a y s i r y o u r a n s w e r i s a l s o c o r r e c t!