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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 :

\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 \dots

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)} .