Question-46720

Question Number 46720 by Necxx last updated on 30/Oct/18

Commented by Necxx last updated on 30/Oct/18

p l e a s e h e l p w i t h n o . 15 c o s i g o t i t

a s

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

Commented by maxmathsup by imad last updated on 30/Oct/18

l e t A = \int \frac{1}{x^{2}} \sqrt{\frac{x - 1}{x + 1}} d x c h a n g e m e n t \frac{x - 1}{x + 1} = t^{2} g i v e x - 1 = t^{2} x + t^{2} \Rightarrow

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

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

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

= 4 a r c t a n (t) - 4 \int \frac{d t}{{(1 + t^{2})}^{2}} c h a n g e m e n t t = t a n θ g i v e

\int \frac{d t}{{(1 + t^{2})}^{2}} = \int \frac{1 + {t a n}^{2} θ}{{(1 + {t a n}^{2} θ)}^{2}} d θ = \int \frac{d θ}{1 + {t a n}^{2} θ} = \int {c o s}^{2} θ d θ =

= \frac{1}{2} \int (1 + c o s (2 θ)) d θ = \frac{θ}{2} + \frac{1}{4} s i n (2 θ) = \frac{a r c t a n (t)}{2} + \frac{1}{4} s i n (2 a r c t a n t) b u t

s i n (2 a r c t a n t) = 2 s i n (a r c t a n t) c o s (a r c t a n t)

= 2 \frac{t}{\sqrt{1 + t^{2}}} . \frac{1}{\sqrt{1 + t^{2}}} = \frac{2 t}{1 + t^{2}} \Rightarrow \int \frac{d t}{{(1 + t^{2})}^{2}} = \frac{a r c t a n (t)}{2} + \frac{t}{2 (1 + t^{2})} \Rightarrow

A = 2 a r c t a n (t) - \frac{2 t}{1 + t^{2}} + c

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

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

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

Commented by Necxx last updated on 30/Oct/18

t h a n k y o u s o m u c h s i r .

T h e i s s u e n o w i s t h a t t h i s a n s w e r

i s n o t i n t h e o p t i o n . D o e s i t m e a n

t h e o p t i o n s a r e a l l w r o n g ?

Commented by maxmathsup by imad last updated on 30/Oct/18

y e s t h e r e i s s o m e e r r o r i n t h e g i v e n a n s w e r s b u t a l w a y s y o u m u s t t r u s t

y o u r s e l f a n d y o u r m e t h o d \dots

Commented by maxmathsup by imad last updated on 30/Oct/18

16) l e t I = \int \frac{d x}{\sqrt{x - x^{2}}} \Rightarrow I = \int \frac{d x}{\sqrt{- (x^{2} - x)}} = \int \frac{d x}{\sqrt{- (x^{2} - 2 \frac{1}{2} x + \frac{1}{4} - \frac{1}{4})}}

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

= \int d t = t + c = a r c s i n (2 x - 1) + c s o o p t i o n (B) i s t h e a n s w e r .

Commented by Necxx last updated on 30/Oct/18

w o w \dots . i s o l v e d t h i s a n d g o t A a s

m y a n s w e r

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