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## Answered question

2021-08-22

Can you indefinite the integral of ${x}^{2}\mathrm{arcsin}\left(x\right)$ please?

### Answer & Explanation

wheezym

Skilled2021-08-23Added 103 answers

indefinite integral of ${x}^{2}.\mathrm{arcsin}\left(x\right)$, using integration by parts.
we use the fact that $\int a\mathrm{sin}\left(x\right)=x.a\mathrm{sin}\left(x\right)+{\left(1-{x}^{2}\right)}^{\frac{1}{2}}$
$\int {x}^{2}.\mathrm{arcsin}\left(x\right)dx={x}^{2}\cdot \left\{x.a\mathrm{sin}\left(x\right)+{\left(1-{x}^{2}\right)}^{\frac{1}{2}}\right\}-\left\{\int 2x.x.a\mathrm{sin}\left(x\right)+2x.{\left(1-{x}^{2}\right)}^{\frac{1}{2}}dx\right\}$
$\int {x}^{2}.\mathrm{arcsin}\left(x\right)dx={x}^{2}\cdot \left\{x.a\mathrm{sin}\left(x\right)+{\left(1-{x}^{2}\right)}^{\frac{1}{2}}\right\}-2\left\{\int {x}^{2}.a\mathrm{sin}\left(x\right)dx\right\}-\left\{\int 2x.{\left(1-{x}^{2}\right)}^{\frac{1}{2}}dx\right\}$
$3\left\{\int {x}^{2}.\mathrm{arcsin}\left(x\right)dx\right\}={x}^{2}\cdot \left\{x.a\mathrm{sin}\left(x\right)+{\left(1-{x}^{2}\right)}^{\frac{1}{2}}\right\}-\left\{\int 2x.{\left(1-{x}^{2}\right)}^{\frac{1}{2}}dx\right\}$
$3\left\{\int {x}^{2}.\mathrm{arcsin}\left(x\right)dx\right\}={x}^{2}\cdot \left\{x.a\mathrm{sin}\left(x\right)+{\left(1-{x}^{2}\right)}^{\frac{1}{2}}\right\}-\left\{-\left(\frac{2}{3}\right){\left(1-{x}^{2}\right)}^{\frac{3}{2}}\right\}$
$3\left\{\int {x}^{2}.\mathrm{arcsin}\left(x\right)dx\right\}={x}^{2}.x.a\mathrm{sin}\left(x\right)+{x}^{2}.{\left(1-{x}^{2}\right)}^{\frac{1}{2}}$

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