Calc1.67

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\int x\arcsin x\,dx\,

First, we must do an integration by parts.

u=\arcsin x\,

du={\frac  {1}{{\sqrt  {1-x^{2}}}}}dx\,

dv=x\,dx\,

v={\frac  {1}{2}}x^{2}\,

\int x\arcsin x\,dx={\frac  {1}{2}}x^{2}\arcsin x-{\frac  {1}{2}}\int {\frac  {x^{2}}{{\sqrt  {1-x^{2}}}}}dx\,

Now, we are at an integral which can be computed using trigonometric substitution.

x=\sin \theta \,

dx=\cos \theta \,d\theta \,

\int x\arcsin x\,dx={\frac  {1}{2}}x^{2}\arcsin x-{\frac  {1}{2}}\int {\frac  {\sin ^{2}\theta \cos \theta }{{\sqrt  {1-\sin ^{2}\theta }}}}d\theta ={\frac  {1}{2}}x^{2}\arcsin x-{\frac  {1}{2}}\int \sin ^{2}\theta \,d\theta \,

Now, let us use the power reduction formula on \sin ^{2}x.

\int x\arcsin x\,dx={\frac  {1}{2}}x^{2}\arcsin x-{\frac  {1}{2}}\int {\frac  {1-\cos 2\theta }{2}}d\theta ={\frac  {1}{2}}x^{2}\arcsin x-{\frac  {1}{4}}\arcsin x+{\frac  {1}{4}}x+C\,

For the back substitution, remember that x=\sin \theta and therefore \theta =\arcsin x.


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