∫_(1/(√2)) ^( 1) ((arcsin x)/x^3 ) dx ? ′ not nice integral ′


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Question Number 127042 by benjo_mathlover last updated on 26/Dec/20

$\int_{1 / \sqrt{2}}^{1} \frac{\arcsin x}{x^{3}} d x ?$ $^{'} n o t n i c e i n t e g r a l^{'}$
Commented by liberty last updated on 26/Dec/20
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	Answered by liberty last updated on 26/Dec/20

	$L = \int_{1 / \sqrt{2}}^{1} \frac{\arcsin x}{x^{3}} d x = {(- \frac{\arcsin x}{2 x^{2}})}_{1 / \sqrt{2}}^{1} + \frac{1}{2} \int_{1 / \sqrt{2}}^{1} \frac{d x}{x^{2} \sqrt{1 - x^{2}}}$ $L = 0 + \frac{1}{2} \int_{1 / \sqrt{2}}^{1} \frac{d x}{x^{2} \sqrt{1 - x^{2}}}$ $[x = \sin h]$ $L = \frac{1}{2} \int_{π / 4}^{π / 2} \frac{\cos h d h}{\sin^{2} h \sqrt{1 - \sin^{2} h}} = \frac{1}{2} \int_{π / 4}^{π / 2} cosec^{2} h d h$ $L = - \frac{1}{2} {[\cot h]}_{π / 4}^{π / 2} = \frac{1}{2}$
	Answered by mathmax by abdo last updated on 27/Dec/20
	$I =∫_(1/( (√2))) ^1 ((arcsinx)/x^3 )dx by parts u^′ =x^(−3) and v=arcsinx ⇒ I =[−(1/(2x^2 ))arcsinx]_(1/( (√2))) ^1 +∫_(1/( (√2))) ^1 (1/(2x^2 ))(dx/( (√(1−x^2 )))) =−(1/2){(π/2)−2×(π/4)}+(1/2)∫_(1/( (√2))) ^1 (dx/(x^2 (√(1−x^2 ))))=(1/2)∫_(1/( (√2))) ^1 (dx/(x^2 (√(1−x^2 )))) changement x=sint give ∫_(1/( (√2))) ^1 (dx/(x^2 (√(1−x^2 ))))=∫_(π/4) ^(π/2) ((cost dt)/(sin^2 t cost)) =2∫_(π/4) ^(π/2) (dt/(1−cos(2t)))=_(2t=α) ∫_(π/2) ^π (dα/(1−cosα)) =_(tan((α/2))=z) =∫_1 ^∞ ((2dz)/((1+z^2 )(1−((1−z^2 )/(1+z^2 ))))) =2∫_1 ^∞ (dz/(1+z^2 −1+z^2 )) =∫_1 ^∞ (dz/z^2 )=[−(1/z)]_1 ^∞ =1 ⇒ ★I =(1/2)★$
	$I = \int_{\frac{1}{\sqrt{2}}}^{1} \frac{arcsinx}{x^{3}} dx by parts u^{^{'}} = x^{- 3} and v = arcsinx \Rightarrow$ $I = {[- \frac{1}{{2 x}^{2}} arcsinx]}_{\frac{1}{\sqrt{2}}}^{1} + \int_{\frac{1}{\sqrt{2}}}^{1} \frac{1}{{2 x}^{2}} \frac{dx}{\sqrt{1 - x^{2}}}$ $= - \frac{1}{2} {\frac{π}{2} - 2 \times \frac{π}{4}} + \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{dx}{x^{2} \sqrt{1 - x^{2}}} = \frac{1}{2} \int_{\frac{1}{\sqrt{2}}}^{1} \frac{dx}{x^{2} \sqrt{1 - x^{2}}}$ $changement x = sint give \int_{\frac{1}{\sqrt{2}}}^{1} \frac{dx}{x^{2} \sqrt{1 - x^{2}}} = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{cost dt}{\sin^{2} t cost}$ $= 2 \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{dt}{1 - \cos (2 t)} =_{2 t = α} \int_{\frac{π}{2}}^{π} \frac{d α}{1 - \cos α} =_{\tan (\frac{α}{2}) = z}$ $= \int_{1}^{\infty} \frac{2 dz}{(1 + z^{2}) (1 - \frac{1 - z^{2}}{1 + z^{2}})} = 2 \int_{1}^{\infty} \frac{dz}{1 + z^{2} - 1 + z^{2}} = \int_{1}^{\infty} \frac{dz}{z^{2}} = {[- \frac{1}{z}]}_{1}^{\infty} = 1 \Rightarrow$ $★ I = \frac{1}{2} ★$
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