Question #76bf7

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Question #76bf7

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Answer 1 · 2017-09-09T21:02:44

$2 x^{3} {sin}^{- 1} x - (\frac{2}{3}) \sqrt{1 - x^{2}} (x^{2} + 2)$ $2x^3sin^-1x - (2/3)sqrt(1-x^2)(x^2+2)$

Explanation:

integration by parts formula:
$\int u \frac{d v}{d x}$ $intu(dv)/dx$ = $uv-int(vu')dx$

So pick something you can integrate (v), and something you can differentiate (u)
$u= sin^-1 x :. u' = 1/sqrt(1 - x^2$
$v=2x^3$ because $v' = 6x^2$

then plug into the formula:

$sin^-1x*2x^3 - int(2x^3)/sqrt(1-x^2)dx$

$(2x^3)/sqrt(1-x^2)$ is a product - can also write as $2x^3*(1-x^2)^(-1/2)$ so will need to use integration by parts again. This comes out as $-(2/3)sqrt(1-x^2)(x^2+2)$

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Question #76bf7

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