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Question Number 188889 by mnjuly1970 last updated on 08/Mar/23

I f, y = \frac{A r c s i n (\sqrt{x})}{\sqrt{x (1 - x)}} \Rightarrow

y^{'} . p (x) + y . q (x) = 1

f i n d, \int_{0}^{1} p (x) . q (x) d x = ?

p, q a r e t w o p l l y n o m i l s \dots

Answered by qaz last updated on 08/Mar/23

y = \frac{\arcsin \sqrt{x}}{\sqrt{x (1 - x)}} \Rightarrow y^{'} = \frac{\frac{1}{2} - \frac{1 - 2 x}{2 \sqrt{x (1 - x)}} \arcsin \sqrt{x}}{x (1 - x)}

= \frac{1 - (1 - 2 x) y}{2 x (1 - x)}

\Rightarrow 2 x (1 - x) y^{'} + (1 - 2 x) y = 1

\Rightarrow p (x) = 2 x (1 - x) q (x) = 1 - 2 x

\Rightarrow \int_{0}^{1} p (x) q (x) d x = \int_{0}^{1} 2 x (1 - x) (1 - 2 x) d x = 0

Commented by mnjuly1970 last updated on 08/Mar/23

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