You need to use the following substitution, such that:
`x = sin t => dx = cos t dt`
`1 - x^2 = 1 - sin^2 t = cos^2 t`
Changing the variable, yields:
`int_0^1 x^3*sqrt(1 - x^2)dx = int_(t_1)^(t_2) sin^3 t*sqrt (cos^2 t)*cos t dt`
`int_(t_1)^(t_2) sin^3 t*sqrt (cos^2 t)*cos t dt = int_(t_1)^(t_2) sin^3 t*cos t*cos t dt `
`int_(t_1)^(t_2) sin^3 t*cos^2 t dt = int_(t_1)^(t_2) sin^2 t*cos^2 t*sin t dt`
`int_(t_1)^(t_2) sin^2 t*cos^2 t*sin tdt = int_(t_1)^(t_2) (1 - cos^2 t)*cos^2 t*sin tdt`
You need to make the next substitution, such that:
`cos t = u => -sin t dt = du`
`int_(u_1)^(u_2) (1 - u^2)*u^2*(-du) =- int_(u_1)^(u_2)u^2 du + int_(u_1)^(u_2)u^4du`
`int_(u_1)^(u_2) (1 - u^2)*u^2*(-du) =(-(cos t)^3/3 + (cos t)^5/5)|_(t_1)^(t_2)`
Since `x = sin t => t = arcsin x`
`int_0^1 x^3*sqrt(1 - x^2)dx = (-(cos (arcsin x))^3/3 + (cos (arcsin x)^5/5)|_(0)^(1)`
`int_0^1 x^3*sqrt(1 - x^2)dx = (-(cos (arcsin 1))^3/3 + (cos (arcsin 1)^5/5 + (cos (arcsin 0))^3/3 - (cos (arcsin 0))^5/5)`
`int_0^1 x^3*sqrt(1 - x^2)dx = -(cos (pi/2))^3/3 + (cos(pi/2))^5/5 + (cos 0)^3/3 - (cos 0)^5/5`
`int_0^1 x^3*sqrt(1 - x^2)dx = 1/3 - 1/5`
`int_0^1 x^3*sqrt(1 - x^2)dx = 2/15`
Hence, evaluating the definite integral yields `int_0^1 x^3*sqrt(1 - x^2)dx = 2/15.`
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