Evaluate ∫_0 ^1 (x^2 /(√(1+x^3 )))dx and given that I_(n ) =∫_0 ^1 x^n (1+x^3 )^(−(1/2)) dx show that (2n−1)I_n =2(√2)−2(n−1) for n≥3. Hence evaluate I_8 , I_7 and I


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Question Number 89273 by Ar Brandon last updated on 16/Apr/20

$E v a l u a t e \int_{0}^{1} \frac{x^{2}}{\sqrt{1 + x^{3}}} d x a n d g i v e n t h a t I_{n} = \int_{0}^{1} x^{n} {(1 + x^{3})}^{- \frac{1}{2}} d x$ $s h o w t h a t (2 n - 1) I_{n} = 2 \sqrt{2} - 2 (n - 1) f o r n ⩾ 3 .$ $H e n c e e v a l u a t e I_{8}, I_{7} a n d I_{6}$
	Answered by 675480065 last updated on 17/Apr/20
	$evaluating the intergral: ∫_0 ^1 (x^2 /(√(1+x^3 )))dx=J Let u=x^3 +1 ⇒du=3x^2 dx ⇒ x^2 dx=(1/3)du. substitute above we get... J=(1/3)∫_0 ^1 u^(−(1/2)) du =(1/6)(√u), but u=(√(x^3 +1)). J=(1/6)((√(√(x^3 +1))))]_0 ^(1 ) =(1/6){(√(√2))−1} Also: I_n =∫_0 ^1 x^n (1+x^3 )^((−1)/2) dx. let u=(1+x^3 )^(−(1/2)) dv=x^n dx ⇒(du/dx)=−((3x^2 )/2)(1+x^3 )^(−(3/2)) v=(x^(n+1) /((n+1))).$
	$evaluating the intergral :$ $\int_{0}^{1} \frac{x^{2}}{\sqrt{1 + x^{3}}} dx = J$ $Let u = x^{3} + 1$ $\Rightarrow du = {3 x}^{2} dx \Rightarrow x^{2} dx = \frac{1}{3} du .$ $substitute above we get . . .$ $J = \frac{1}{3} \int_{0}^{1} u^{- \frac{1}{2}} du = \frac{1}{6} \sqrt{u}, but u = \sqrt{x^{3} + 1} .$ ${J = \frac{1}{6} (\sqrt{\sqrt{x^{3} + 1}})]}_{0}^{1} = \frac{1}{6} {\sqrt{\sqrt{2}} - 1}$ $Also :$ $I_{n} = \int_{0}^{1} x^{n} {(1 + x^{3})}^{\frac{- 1}{2}} dx .$ $let u = {(1 + x^{3})}^{- \frac{1}{2}} dv = x^{n} dx$ $\Rightarrow \frac{du}{dx} = - \frac{{3 x}^{2}}{2} {(1 + x^{3})}^{- \frac{3}{2}} v = \frac{x^{n + 1}}{(n + 1)} .$
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