# Can you indefinite the integral of x^2.arcsin(x) please?

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

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wheezym
indefinite integral of $$\displaystyle{x}^{{2}}.{\arcsin{{\left({x}\right)}}}$$, using integration by parts.
we use the fact that $$\displaystyle\int{a}{\sin{{\left({x}\right)}}}={x}.{a}{\sin{{\left({x}\right)}}}+{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}$$
$$\displaystyle\int{x}^{{2}}.{\arcsin{{\left({x}\right)}}}{\left.{d}{x}\right.}={x}^{{2}}\cdot{\left\lbrace{x}.{a}{\sin{{\left({x}\right)}}}+{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}\right\rbrace}-{\left\lbrace\int{2}{x}.{x}.{a}{\sin{{\left({x}\right)}}}+{2}{x}.{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}{\left.{d}{x}\right.}\right\rbrace}$$
$$\displaystyle\int{x}^{{2}}.{\arcsin{{\left({x}\right)}}}{\left.{d}{x}\right.}={x}^{{2}}\cdot{\left\lbrace{x}.{a}{\sin{{\left({x}\right)}}}+{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}\right\rbrace}-{2}{\left\lbrace\int{x}^{{2}}.{a}{\sin{{\left({x}\right)}}}{\left.{d}{x}\right.}\right\rbrace}-{\left\lbrace\int{2}{x}.{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}{\left.{d}{x}\right.}\right\rbrace}$$
$$\displaystyle{3}{\left\lbrace\int{x}^{{2}}.{\arcsin{{\left({x}\right)}}}{\left.{d}{x}\right.}\right\rbrace}={x}^{{2}}\cdot{\left\lbrace{x}.{a}{\sin{{\left({x}\right)}}}+{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}\right\rbrace}-{\left\lbrace\int{2}{x}.{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}{\left.{d}{x}\right.}\right\rbrace}$$
$$\displaystyle{3}{\left\lbrace\int{x}^{{2}}.{\arcsin{{\left({x}\right)}}}{\left.{d}{x}\right.}\right\rbrace}={x}^{{2}}\cdot{\left\lbrace{x}.{a}{\sin{{\left({x}\right)}}}+{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}\right\rbrace}-{\left\lbrace-{\left(\frac{{2}}{{3}}\right)}{\left({1}-{x}^{{2}}\right)}^{{\frac{{3}}{{2}}}}\right\rbrace}$$
$$\displaystyle{3}{\left\lbrace\int{x}^{{2}}.{\arcsin{{\left({x}\right)}}}{\left.{d}{x}\right.}\right\rbrace}={x}^{{2}}.{x}.{a}{\sin{{\left({x}\right)}}}+{x}^{{2}}.{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}+{\left(\frac{{2}}{{3}}\right)}{\left({1}-{x}^{{2}}\right)}^{{\frac{{3}}{{2}}}}$$
$$\displaystyle{3}{\left\lbrace\int{x}^{{2}}.{\arcsin{{\left({x}\right)}}}{\left.{d}{x}\right.}\right\rbrace}={x}^{{3}}.{a}{\sin{{\left({x}\right)}}}+{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}{\left\lbrace{x}^{{2}}+{\left(\frac{{2}}{{3}}\right)}{\left({1}-{x}^{{2}}\right)}\right\rbrace}$$
$$\displaystyle\int{x}^{{2}}.{\arcsin{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{3}}\right)}.{x}^{{3}}.{a}{\sin{{\left({x}\right)}}}+{\left(\frac{{1}}{{9}}\right)}.{\left\lbrace{3}{x}^{{2}}+{2}{\left({1}-{x}^{{2}}\right)}\right\rbrace}.{\left({1}-{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}$$
$$\displaystyle\int{x}^{{2}}.{\arcsin{{\left({x}\right)}}}{\left.{d}{x}\right.}={\left(\frac{{1}}{{3}}\right)}.{x}^{{3}}.{a}{\sin{{\left({x}\right)}}}+{\left(\frac{{1}}{{9}}\right)}.{\left\lbrace{x}^{{2}}+{2}{x}^{{2}}\right)}^{{\frac{{1}}{{2}}}}$$
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