Question

Solve the equation-
$\int x^4\sqrt{1-x^2}dx$

Answer 1

$\int x^4\sqrt{1-x^2}dx$

Let $x=\sin u$ then $dx=\cos udu$

$=\int {\sin }^{4}u\sqrt{1-{\sin }^{2}u}\cos udu$

$=\int {\sin }^{4}u\sqrt{{\cos }^{2}u}\cos udu$

$=\int {\sin }^{4}u{\cos }^{2}udu$

Applying the integral reduction formula,

$\int {\sin }^{n}x{\cos }^{2}xdx=\frac{-{\sin }^{n-1}x{\cos }^{3}x}{n+2}+\frac{n-1}{n+2}\int {\sin }^{n-2}x{\cos }^{2}xdx$

$=\frac{-{\sin }^{3}u{\cos }^{3}u}{6}+\frac{3}{6}\int {\sin }^{2}u{\cos }^{2}udu$

$=\frac{-{\sin }^{3}u{\cos }^{3}u}{6}+\frac{1}{2}\int \frac{1-\cos 4u}{8}du$

$=\frac{-{\sin }^{3}u{\cos }^{3}u}{6}+\frac{1}{2}\left(\int \frac{1}{8}du-\int \frac{\cos 4u}{8}du\right)$

$=\frac{-{\sin }^{3}u{\cos }^{3}u}{6}+\frac{1}{16}(u-\frac{1}{4}\sin 4u)$

$=\frac{-{\sin }^3\left({\sin }^{-1}x\right){\cos }^3\left({\sin }^{-1}x\right)}{6}+\frac{1}{16}(\left({\sin }^{-1}x\right)-\frac{1}{4}\sin 4\left({\sin }^{-1}x\right))+c$

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