sin-1-x-2-dx-

Question Number 93045 by john santu last updated on 10/May/20

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

Commented by mathmax by abdo last updated on 10/May/20

I = \int a r c s i n (\sqrt{\frac{x}{2}}) d x c h a n g e m e n t \sqrt{\frac{x}{2}} = t g i v e \frac{x}{2} = t^{2} \Rightarrow x = 2 t^{2}

I = \int a r c s i n (t) (4 t) d t = 4 \int t a r c s i n (t) d t a n d b y p a r t s

I = 4 {\frac{t^{2}}{2} a r c s i n t - \int \frac{t^{2}}{2} \frac{d t}{\sqrt{1 - t^{2}}}} = 2 t^{2} a r c s i n t - 2 \int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t

\int \frac{t^{2}}{\sqrt{1 - t^{2}}} d t =_{t = s i n α} \int \frac{{s i n}^{2} α}{c o s α} c o s α d α = \int {s i n}^{2} α d α

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

= \frac{α}{2} - \frac{1}{2} s i n α c o s α + c = \frac{1}{2} a r c s i n t - \frac{1}{2} t \sqrt{1 - t^{2}} + c

= \frac{1}{2} a r c s i n (\sqrt{\frac{x}{2}}) - \frac{1}{2} \sqrt{\frac{x}{2}} \sqrt{1 - \frac{x}{2}} + c \Rightarrow

I = x a r c s i n (\sqrt{\frac{x}{2}}) - a r c s i n (\sqrt{\frac{x}{2}}) + \sqrt{\frac{x}{2}} \sqrt{1 - \frac{x}{2}} + c

Commented by john santu last updated on 11/May/20

yes . i got

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

Answered by TANMAY PANACEA … last updated on 10/May/20

x = 2 {s i n}^{2} t

d x = 2 \times s i n 2 t \times d t

\int t \times 2 s i n 2 t d t

\frac{I}{2} = \int t . s i n 2 t d t

= t \int s i n 2 t d t - \int [\frac{d t}{d t} \int s i n 2 t d t] d t

= t \times \frac{c o s 2 t}{- 2} - \int 1 \times \frac{c o s 2 t}{- 2} d t

= \frac{- t c o s 2 t}{2} + \frac{1}{2} \times \frac{s i n 2 t}{2} + C

= \frac{- t}{2} \times (1 - 2 {s i n}^{2} t) + \frac{1}{4} \times 2 s i n t \times c o s t + c

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

s o I = - {s i n}^{- 1} (\sqrt{\frac{x}{2}}) (1 - x) + \sqrt{\frac{x}{2}} {(1 - \frac{x}{2})}^{\frac{1}{2}} + C_{1}