Blog: `int (arcsin(x))^2 dx` Evaluate the integral

Thursday, October 12, 2017

$\displaystyle\int{{\left({\arcsin{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}$ Evaluate the integral

You need to use integration by parts, such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}\Rightarrow{d}{u}=\frac{{{2}{\arcsin{{x}}}}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{d}{v}={1}\Rightarrow{v}={x}$

$\displaystyle\int{\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}={x}\cdot{\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}-{2}\int{\arcsin{{x}}}\cdot\frac{{x}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}{\left.{d}{x}\right.}$

You need to solve the integral $\displaystyle\int{\arcsin{{x}}}\cdot{\left(\frac{{x}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}\right)}{\left.{d}{x}\right.}$ using substitution $\displaystyle{\arcsin{{x}}}={t}$ , such that:

$\displaystyle{\arcsin{{x}}}={t}\Rightarrow\frac{{{\left.{d}{x}\right.}}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}={\left.{d}{t}\right.}$

$\displaystyle\int{\arcsin{{x}}}\cdot{\left(\frac{{x}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}\right)}{\left.{d}{x}\right.}=\int{t}\cdot{\sin{{t}}}{\left.{d}{t}\right.}$

You need to use...

You need to use integration by parts, such that:

$\displaystyle\int{u}{d}{v}={u}{v}-\int{v}{d}{u}$

$\displaystyle{u}={\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}\Rightarrow{d}{u}=\frac{{{2}{\arcsin{{x}}}}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}{\left.{d}{x}\right.}$

$\displaystyle{d}{v}={1}\Rightarrow{v}={x}$

$\displaystyle{\arcsin{{x}}}={t}\Rightarrow\frac{{{\left.{d}{x}\right.}}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}={\left.{d}{t}\right.}$

$\displaystyle\int{\arcsin{{x}}}\cdot{\left(\frac{{x}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}\right)}{\left.{d}{x}\right.}=\int{t}\cdot{\sin{{t}}}{\left.{d}{t}\right.}$

You need to use integration by parts to solve $\displaystyle\int{t}\cdot{\sin{{t}}}{\left.{d}{t}\right.}$ , such that:

$\displaystyle{u}={t}\Rightarrow{d}{u}={\left.{d}{t}\right.}$

$\displaystyle{d}{v}={\sin{{t}}}\Rightarrow{v}=-{\cos{{t}}}$

$\displaystyle\int{t}\cdot{\sin{{t}}}{\left.{d}{t}\right.}=-{t}\cdot{\cos{{t}}}+\int{\cos{{t}}}{\left.{d}{t}\right.}$

$\displaystyle\int{t}\cdot{\sin{{t}}}{\left.{d}{t}\right.}=-{t}\cdot{\cos{{t}}}+{\sin{{t}}}+{c}$

Replacing back the variable yields:

$\displaystyle\int{\arcsin{{x}}}\cdot{\left(\frac{{x}}{{\sqrt{{{1}-{{x}}^{{2}}}}}}\right)}{\left.{d}{x}\right.}=-{\arcsin{{x}}}\cdot{\cos{{\left({\arcsin{{x}}}\right)}}}+{x}+{c}$

$\displaystyle\int{\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}={x}\cdot{\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}-{2}{\left(-{\arcsin{{x}}}\cdot{\cos{{\left({\arcsin{{x}}}\right)}}}+{x}\right)}+{C}$

$\displaystyle\int{\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}={x}\cdot{\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}+{2}{\arcsin{{x}}}\cdot{\cos{{\left({\arcsin{{x}}}\right)}}}-{2}{x}+{c}$

Hence, evaluating the indefinite integral, using parts and substitution, yields $\displaystyle\int{\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}{\left.{d}{x}\right.}={x}\cdot{\left({{\arcsin}}^{{2}}{\left({x}\right)}\right)}+{2}{\arcsin{{x}}}\cdot{\cos{{\left({\arcsin{{x}}}\right)}}}-{2}{x}+{c}.$

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Thursday, October 12, 2017

$\displaystyle\int{{\left({\arcsin{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}$ Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?

Thursday, October 12, 2017

Evaluate the integral

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What is the Exposition, Rising Action, Climax, and Falling Action of &quot;One Thousand Dollars&quot;?

$\displaystyle\int{{\left({\arcsin{{\left({x}\right)}}}\right)}}^{{2}}{\left.{d}{x}\right.}$ Evaluate the integral

What is the Exposition, Rising Action, Climax, and Falling Action of "One Thousand Dollars"?